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线段树专题
①POJ3667
题目链接：http://poj.org/problem?id=3667
思路：pushup为合并，向上更新。pushdown为向下更新，lazy存放待更新的操作，1为置满，0为无操作，-1为置空。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 5e4 + 5;
struct node
{
int l,r,lazy,ls,rs,ms;
}tree[N<<2];
void pushup(int i)
{
int mid = (tree[i].l + tree[i].r) >> 1;
if(tree[i*2].ls == mid - tree[i].l + 1)
tree[i].ls = tree[i*2].ls + tree[i*2 + 1].ls;
else
tree[i].ls = tree[i*2].ls;
if(tree[i*2 + 1].rs == tree[i].r - mid)
tree[i].rs = tree[i*2 + 1].rs + tree[i*2].rs;
else
tree[i].rs = tree[i*2 + 1].rs;
tree[i].ms = max(max(tree[i*2].ms,tree[i*2 + 1].ms),tree[i*2].rs + tree[i*2 + 1].ls);
}
void pushdown(int i)
{
if(tree[i].lazy)
{
tree[i*2].lazy = tree[i*2 + 1].lazy = tree[i].lazy;
if(tree[i].lazy == 1)
{
tree[i*2].ls = tree[i*2].rs = tree[i*2].ms = 0;
tree[i*2 + 1].ls = tree[i*2 + 1].rs = tree[i*2 + 1].ms = 0;
}
else
{
tree[i*2].ls = tree[i*2].rs = tree[i*2].ms = tree[i*2].r - tree[i*2].l + 1;
tree[i*2 + 1].ls = tree[i*2 + 1].rs = tree[i*2 + 1].ms = tree[i*2 + 1].r - tree[i*2 + 1].l + 1;
}
tree[i].lazy = 0;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].lazy = 0;
tree[i].ls = tree[i].rs = tree[i].ms = r - l + 1;
if(l == r)
return;
int mid = (l + r) >> 1;
built(i*2,l,mid);
built(i*2 + 1,mid + 1,r);
}
void update(int i,int l,int r,int c)
{
if(tree[i].l >= l && tree[i].r <= r)
{
tree[i].lazy = c;
if(c == 1)
tree[i].ls = tree[i].rs = tree[i].ms = 0;
else
tree[i].ls = tree[i].rs = tree[i].ms = tree[i].r - tree[i].l + 1;
return;
}
if(tree[i].r < l || tree[i].l > r)
return;
pushdown(i);
update(i*2,l,r,c);
update(i*2 + 1,l,r,c);
pushup(i);
}
int query(int i,int len)
{
pushdown(i);
if(tree[i].ls >= len)
return tree[i].l;
if(tree[i*2].ms >= len)
return query(i*2,len);
if(tree[i*2].rs + tree[i*2 + 1].ls >= len)
{
int mid = (tree[i].l + tree[i].r) >> 1;
return mid - tree[i*2].rs + 1;
}
return query(i*2 + 1,len);
}
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
built(1,1,n);
while(m--)
{
int c;
scanf("%d",&c);
if(c == 1)
{
int x;
scanf("%d",&x);
if(tree[1].ms < x)
{
puts("0");
continue;
}
int ans = query(1,x);
printf("%d\n",ans);
update(1,ans,ans + x - 1,1);
}
else
{
int x,y;
scanf("%d %d",&x,&y);
update(1,x,x + y - 1,-1);
}
}
}
return 0;
}
②POJ3225
题目链接:http://poj.org/problem?id=3225
#include<cstdio>
#define le(i) i<<1
#define ri(i) i<<1|1
using namespace std;
const int N = 65535 << 1;
struct node
{
int l,r,lazy,cov;//lazy表示操作：2为互换，1为置满，0为置空，-1为无操作。cov表示状态：1为全包含，0为全不包含，-1为其它
}tree[(N<<2)+5];
int s[(N<<1)+5];
void pushup(int i)
{
if(tree[le(i)].cov == tree[ri(i)].cov)
tree[i].cov = tree[le(i)].cov;
else
tree[i].cov = -1;
}
void pushdown(int i)
{
if(tree[i].lazy != -1)
{
if(tree[le(i)].lazy != -1 && tree[i].lazy == 2)
{
if(tree[le(i)].lazy == 2)
tree[le(i)].lazy = -1;
else
tree[le(i)].lazy ^= 1;
}
else
tree[le(i)].lazy = tree[i].lazy;
if(tree[ri(i)].lazy != -1 && tree[i].lazy == 2)
{
if(tree[ri(i)].lazy == 2)
tree[ri(i)].lazy = -1;
else
tree[ri(i)].lazy ^= 1;
}
else
tree[ri(i)].lazy = tree[i].lazy;
if(tree[i].lazy == 2)
{
if(tree[le(i)].cov != -1)
tree[le(i)].cov ^= 1;
if(tree[ri(i)].cov != -1)
tree[ri(i)].cov ^= 1;
}
else
tree[le(i)].cov = tree[ri(i)].cov = tree[i].lazy;
tree[i].lazy = -1;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].lazy = -1;
tree[i].cov = 0;
if(l == r)
return;
int mid = (l + r) >> 1;
built(le(i),l,mid);
built(ri(i),mid + 1,r);
}
void update(int i,int l,int r,int op)//op:0为置空,1为置满，2为互换
{
if(r < l)
return;
if(tree[i].l >= l && tree[i].r <= r)
{
if(tree[i].lazy != -1 && op == 2)
{
if(tree[i].lazy == 2)
tree[i].lazy = -1;
else
tree[i].lazy ^= 1;
}
else
tree[i].lazy = op;
if(op == 2)
{
if(tree[i].cov != -1)
tree[i].cov ^= 1;
}
else
tree[i].cov = op;
return;
}
if(tree[i].l > r|| tree[i].r < l)
return;
pushdown(i);
update(le(i),l,r,op);
update(ri(i),l,r,op);
pushup(i);
}
void query(int i)
{
if(!tree[i].cov)
return;
if(tree[i].cov == 1)
{
for(int k = tree[i].l; k <= tree[i].r; k++)
s[k] = 1;
return;
}
else
{
pushdown(i);
query(le(i));
query(ri(i));
}
}
int main()
{
char c,l0,r0;
int l,r;
built(1,0,N);
while(scanf(" %c %c%d,%d %c",&c,&l0,&l,&r,&r0) != EOF)
{
if(l0 == ‘(‘)
l = l<<1|1;
else
l <<= 1;
if(r0 == ‘)‘)
r = (r<<1) - 1;
else
r <<= 1;
if(c == ‘U‘)
update(1,l,r,1);
else if(c == ‘I‘)
{
update(1,0,l-1,0);
update(1,r+1,N,0);
}
else if(c == ‘D‘)
update(1,l,r,0);
else if(c == ‘C‘)
{
update(1,0,l-1,0);
update(1,r+1,N,0);
update(1,l,r,2);
}
else
update(1,l,r,2);
}
query(1);
int a = -1,flag = 0;
for(int i = 0; i <= N + 1; i++)
{
if(s[i] && a == -1)
a = i,flag = 1;
if(!s[i] && a != -1)
{
if(a&1)
printf("(%d,",a>>1);
else
printf("[%d,",a>>1);
if(i&1)
printf("%d] ",i>>1);
else
printf("%d) ",i>>1);
a = -1;
}
}
if(!flag)
printf("empty set");
printf("\n");
return 0;
}
③POJ5023
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5023
由于颜色只有30种，所以可以用位存储，网上看到的好神奇，我却只能想到set
#include<bits/stdc++.h>
#define le i<<1
#define ri i<<1|1
using namespace std;
const int N = 1e6 + 5;
int ans;
struct node
{
int l,r,lazy,color;
}tree[N<<2];
void pushup(int i)
{
tree[i].color = tree[le].color | tree[ri].color;
}
void pushdown(int i)
{
if(tree[i].lazy)
{
tree[le].lazy = tree[ri].lazy = tree[i].lazy;
tree[le].color = tree[ri].color = 1<<tree[i].lazy;
tree[i].lazy = 0;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].lazy = 0;
tree[i].color = 4;
if(l == r)
return;
int mid = (l + r) >> 1;
built(le,l,mid);
built(ri,mid + 1,r);
}
void update(int i,int l,int r,int c)
{
if(tree[i].l >= l && tree[i].r <= r)
{
tree[i].lazy = c;
tree[i].color = 1<<c;
return;
}
if(tree[i].l > r || tree[i].r < l)
return;
pushdown(i);
update(le,l,r,c);
update(ri,l,r,c);
pushup(i);
}
void query(int i,int l,int r)
{
if(tree[i].l >= l && tree[i].r <= r)
{
ans |= tree[i].color;
return;
}
if(tree[i].l > r || tree[i].r < l)
return;
pushdown(i);
query(le,l,r);
query(ri,l,r);
}
int main()
{
int n,m,c,l,r;
char com;
while(scanf("%d %d",&n,&m))
{
if(!n && !m)
break;
built(1,1,n);
while(m--)
{
scanf(" %c",&com);
if(com == ‘P‘)
{
scanf("%d %d %d",&l,&r,&c);
update(1,l,r,c);
}
else
{
scanf("%d %d",&l,&r);
query(1,l,r);
int flag = 0;
for(int i = 1; i <= 30; i++)
{
ans >>= 1;
if(ans & 1)
{
if(!flag)
printf("%d",i), flag = 1;
else
printf(" %d",i);
}
}
puts("");
}
ans = 0;
}
}
return 0;
}
④HDU4578
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4578
思路：对于操作1、2，设k为增加倍数，b为增加数量，初始状态k = 1，b = 0
有递推公式sum1 = ∑(kx + b) = k*sum1 + b*len
sum2 = ∑(kx + b)2 = ∑(k2x2 + 2kxb + b2) = k2 * sum2 + 2*k*sum1*b + b2 * len
sum3 = ∑(kx + b)3 = ∑(k3x3 + 3k2x2b + 3kxb2 + b3) = k3 * sum3 + 3 * k2 * sum2 * b + 3 * k * sum1 * b2 + b3 * len
对于操作3，会覆盖之前的操作，之前的操作应清零
该死的宏定义
#include<bits/stdc++.h>
#define N 100005
#define M 10007
#define le i<<1
#define ri i<<1|1
using namespace std;
struct node
{
int l,r,lazy1,lazy2,lazy3,sum1,sum2,sum3;
int len()
{
return (r - l + 1) % M;
}
}tree[N<<2];
void add_mul(int i,int k,int b)
{
int s1 = tree[i].sum1 % M;
int s2 = tree[i].sum2 % M;
int s3 = tree[i].sum3 % M;
int len = tree[i].len() % M;
tree[i].lazy1 = (k % M * tree[i].lazy1 % M + b % M) % M;
tree[i].lazy2 = k % M * tree[i].lazy2 % M;
tree[i].sum1 = (k % M * s1 % M + b % M * len % M) % M;
tree[i].sum2 = (k % M * k % M * s2 % M + 2 * k % M * s1 % M * b % M + b % M * b % M * len % M) % M;
tree[i].sum3 = (k % M * k % M * k % M * s3 % M + 3 * k % M * k % M * s2 % M * b % M + 3 * k % M * s1 % M * b % M * b % M + b % M * b % M * b % M * len % M) % M;
}
void equ(int i,int c)
{
tree[i].lazy1 = 0;
tree[i].lazy2 = 1;
tree[i].lazy3 = c;
tree[i].sum1 = c * tree[i].len() % M;
tree[i].sum2 = c * c % M * tree[i].len() % M;
tree[i].sum3 = c * c % M * c % M * tree[i].len() % M;
}
void pushup(int i)
{
tree[i].sum1 = (tree[le].sum1 + tree[ri].sum1) % M;
tree[i].sum2 = (tree[le].sum2 + tree[ri].sum2) % M;
tree[i].sum3 = (tree[le].sum3 + tree[ri].sum3) % M;
}
void pushdown(int i)
{
if(tree[i].lazy3)
{
equ(le,tree[i].lazy3);
equ(ri,tree[i].lazy3);
tree[i].lazy3 = 0;
}
if(tree[i].lazy1 != 0 || tree[i].lazy2 != 1)
{
add_mul(le,tree[i].lazy2,tree[i].lazy1);
add_mul(ri,tree[i].lazy2,tree[i].lazy1);
tree[i].lazy1 = 0;
tree[i].lazy2 = 1;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].lazy1 = 0;
tree[i].lazy2 = 1;
tree[i].lazy3 = 0;
tree[i].sum1 = tree[i].sum2 = tree[i].sum3 = 0;
if(l == r)
return;
int mid = (l + r) >> 1;
built(le,l,mid);
built(ri,mid + 1,r);
}
void update(int i,int l,int r,int op,int c)
{
if(tree[i].l >= l && tree[i].r <= r)
{
if(op == 1)
add_mul(i,1,c);
else if(op == 2)
add_mul(i,c,0);
else
equ(i,c);
return;
}
if(tree[i].l > r || tree[i].r < l)
return;
pushdown(i);
update(le,l,r,op,c);
update(ri,l,r,op,c);
pushup(i);
}
int query(int i,int l,int r,int c)
{
if(tree[i].l >= l && tree[i].r <= r)
{
if(c == 1)
return tree[i].sum1 % M;
if(c == 2)
return tree[i].sum2 % M;
if(c == 3)
return tree[i].sum3 % M;
}
if(tree[i].l > r || tree[i].r < l)
return 0;
pushdown(i);
return (query(le,l,r,c) + query(ri,l,r,c)) % M;
}
int main()
{
int n,m;
while(scanf("%d %d",&n,&m))
{
if(n == 0 && m == 0)
break;
built(1,1,n);
while(m--)
{
int op,x,y,c;
scanf("%d %d %d %d",&op,&x,&y,&c);
if(op == 4)
printf("%d\n",query(1,x,y,c) % M);
else
update(1,x,y,op,c);
}
}
return 0;
}
⑥HDU3397
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3397
#include<bits/stdc++.h>
#define lson i<<1
#define rson i<<1|1
using namespace std;
const int N = 1e5 + 5;
int str[N];
struct node
{
int l,r;
int lazy;
int num1;
int ls1,rs1,ms1;
int ls0,rs0,ms0;
int len()
{
return r - l + 1;
}
}tree[N<<2];
void change(int i,int op)
{
if(op == 2)
{
swap(tree[i].ls0,tree[i].ls1);
swap(tree[i].rs0,tree[i].rs1);
swap(tree[i].ms0,tree[i].ms1);
tree[i].lazy = 1 - tree[i].lazy;
tree[i].num1 = tree[i].len() - tree[i].num1;
}
else if(op == 1)
{
tree[i].lazy = 1;
tree[i].ls1 = tree[i].rs1 = tree[i].ms1 = tree[i].num1 = tree[i].len();
tree[i].ls0 = tree[i].rs0 = tree[i].ms0 = 0;
}
else
{
tree[i].lazy = 0;
tree[i].ls0 = tree[i].rs0 = tree[i].ms0 = tree[i].len();
tree[i].ls1 = tree[i].rs1 = tree[i].ms1 = tree[i].num1 = 0;
}
}
void pushup(int i)
{
if(tree[lson].ls1 == tree[lson].len())
tree[i].ls1 = tree[lson].ls1 + tree[rson].ls1;
else
tree[i].ls1 = tree[lson].ls1;
if(tree[rson].rs1 == tree[rson].len())
tree[i].rs1 = tree[rson].rs1 + tree[lson].rs1;
else
tree[i].rs1 = tree[rson].rs1;
if(tree[lson].ls0 == tree[lson].len())
tree[i].ls0 = tree[lson].ls0 + tree[rson].ls0;
else
tree[i].ls0 = tree[lson].ls0;
if(tree[rson].rs0 == tree[rson].len())
tree[i].rs0 = tree[rson].rs0 + tree[lson].rs0;
else
tree[i].rs0 = tree[rson].rs0;
tree[i].ms1 = max(max(tree[lson].ms1,tree[rson].ms1),tree[lson].rs1 + tree[rson].ls1);
tree[i].ms0 = max(max(tree[lson].ms0,tree[rson].ms0),tree[lson].rs0 + tree[rson].ls0);
tree[i].num1 = tree[lson].num1 + tree[rson].num1;
}
void pushdown(int i)
{
if(tree[i].lazy != -1)
{
change(lson,tree[i].lazy);
change(rson,tree[i].lazy);
tree[i].lazy = -1;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].lazy = -1;
if(l == r)
{
tree[i].num1 = str[l];
tree[i].ls1 = tree[i].rs1 = tree[i].ms1 = str[l];
tree[i].ls0 = tree[i].rs0 = tree[i].ms0 = str[l]^1;
return;
}
int mid = (l + r) >> 1;
built(lson,l,mid);
built(rson,mid+1,r);
pushup(i);
}
void update(int i,int l,int r,int op)
{
if(tree[i].l >= l && tree[i].r <= r)
{
change(i,op);
return;
}
if(tree[i].l > r || tree[i].r < l)
return;
pushdown(i);
update(lson,l,r,op);
update(rson,l,r,op);
pushup(i);
}
int query(int i,int l,int r,int op)
{
if(tree[i].l >= l && tree[i].r <= r)
{
if(op == 3)
return tree[i].num1;
else
return tree[i].ms1;
}
if(tree[i].l > r || tree[i].r < l)
return 0;
pushdown(i);
if(op == 3)
return query(lson,l,r,op) + query(rson,l,r,op);
else
{
int res1 = min(tree[lson].rs1,tree[lson].r - l + 1) + min(tree[rson].ls1,r - tree[rson].l + 1);
int res2 = max(query(lson,l,r,op),query(rson,l,r,op));
return max(res1,res2);
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d %d",&n,&m);
for(int i = 0; i < n; i++)
scanf("%d",str+i);
built(1,0,n-1);
while(m--)
{
int op,a,b;
scanf("%d %d %d",&op,&a,&b);
if(op <= 2)
update(1,a,b,op);
else
printf("%d\n",query(1,a,b,op));
}
}
return 0;
}
⑦HDU3016（DAG最长路DP + 线段树区间更新）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3016
题意：本来看错了，以为板上都可以跳。其实只能在板的两端垂直下落。
//不得不承认专项练习还是有个弊端，就是知道它归类于线段树就会拼命往线段树想，要是直接上手的话我肯定想不到。
思路：downl，downr分别表示从该板左右两端跳会落到哪块板子上（id），这样就可以利用线段树建立出一个DAG。然后就是经典的最长路DP。
#include<bits/stdc++.h>
#define lson i<<1
#define rson i<<1|1
using namespace std;
const int N = 1e5 + 5;
const int inf = 0x3f3f3f3f;
int dp[N];
struct Board
{
int l,r;
int h,val;
int downl,downr;
bool operator < (Board t)
{
return h < t.h;
}
}board[N];
struct node
{
int l,r;
int cov;
}tree[N<<2];
void pushup(int i)
{
if(tree[lson].cov == tree[rson].cov)
tree[i].cov = tree[lson].cov;
else
tree[i].cov = -1;
}
void pushdown(int i)
{
if(tree[i].cov != -1)
{
tree[lson].cov = tree[rson].cov = tree[i].cov;
tree[i].cov = -1;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].cov = 0;
if(l == r)
return;
int mid = (l + r) >> 1;
built(lson,l,mid);
built(rson,mid+1,r);
}
void update(int i,int l,int r,int id)
{
if(tree[i].l >= l && tree[i].r <= r)
{
tree[i].cov = id;
return;
}
if(tree[i].l > r || tree[i].r < l)
return;
pushdown(i);
update(lson,l,r,id);
update(rson,l,r,id);
pushup(i);
}
int query(int i,int c)
{
if(tree[i].cov != -1)
return tree[i].cov;
int mid = (tree[i].l + tree[i].r) >> 1;
if(c <= mid)
return query(lson,c);
else
return query(rson,c);
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i = 1; i <= n; i++)
scanf("%d %d %d %d",&board[i].h,&board[i].l,&board[i].r,&board[i].val);
sort(board + 1,board + n + 1);
built(1,1,100000);
for(int i = 1; i <= n; i++)
{
board[i].downl = query(1,board[i].l);
board[i].downr = query(1,board[i].r);
update(1,board[i].l,board[i].r,i);
}
memset(dp,-inf,sizeof(dp));
dp[n] = 100 + board[n].val;
for(int i = n; i >= 1; i--)
{
if(dp[i] < 0)
dp[i] = -inf;
dp[board[i].downl] = max(dp[board[i].downl], dp[i] + board[board[i].downl].val);
dp[board[i].downr] = max(dp[board[i].downr], dp[i] + board[board[i].downr].val);
}
if(dp[0] <= 0)
puts("-1");
else
printf("%d\n",dp[0]);
}
return 0;
}
⑧POJ3577
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3577
#include<bits/stdc++.h>
#define lson i<<1
#define rson i<<1|1
using namespace std;
const int N = 1e6 + 5;
bool yes[100005];
struct node
{
int l,r;
int lazy,maxm;
}tree[N<<2];
void pushdown(int i)
{
if(tree[i].lazy)
{
tree[lson].lazy += tree[i].lazy;
tree[rson].lazy += tree[i].lazy;
tree[lson].maxm += tree[i].lazy;
tree[rson].maxm += tree[i].lazy;
tree[i].lazy = 0;
}
}
void built(int i,int l,int r)
{
tree[i].l = l;
tree[i].r = r;
tree[i].lazy = tree[i].maxm = 0;
if(l == r)
return;
int mid = (l + r) >> 1;
built(lson,l,mid);
built(rson,mid+1,r);
}
void update(int i,int l,int r)
{
if(tree[i].l >= l && tree[i].r <= r)
{
tree[i].lazy++;
tree[i].maxm++;
return;
}
if(tree[i].l > r || tree[i].r < l)
return;
pushdown(i);
update(lson,l,r);
update(rson,l,r);
tree[i].maxm = max(tree[lson].maxm,tree[rson].maxm);
}
int query(int i,int l,int r)
{
if(tree[i].l >= l && tree[i].r <= r)
return tree[i].maxm;
if(tree[i].l > r || tree[i].r < l)
return 0;
pushdown(i);
return max(query(lson,l,r),query(rson,l,r));
}
int main()
{
int T,cas = 1,k,Q,a,b;
scanf("%d",&T);
while(T--)
{
memset(yes,0,sizeof(yes));
scanf("%d %d",&k,&Q);
built(1,1,1000000);
for(int i = 1; i <= Q; i++)
{
scanf("%d %d",&a,&b);
if(query(1,a,b-1) < k)
{
yes[i] = 1;
update(1,a,b-1);
}
}
printf("Case %d:\n",cas++);
for(int i = 1; i <= Q; i++)
{
if(yes[i])
printf("%d ",i);
}
printf("\n\n");
}
return 0;
}
⑨HDU5316（单点更新，区间合并）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5316
题意：给出一个长度为n的序列，有两种操作，op = 0时求出区间[a,b]中奇偶相间（不含空）的子序列的和的最大值，op = 1时把a改成b
思路：oo表示该节点区间[L,R]上以奇开头奇结尾的子序列的和的最大值，oe，eo，ee同理......
自己之前的查询部分写的有点赘，TLE了，网上看了别人的写了下，发现之前的几题也可以简化下代码量的。
#include<bits/stdc++.h>
#define lson i<<1
#define rson i<<1|1
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
const ll inf = 1e18;
ll MAX(ll a,ll b,ll c,ll d)
{
return max(max(a,b),max(c,d));
}
struct node
{
ll oo,oe,eo,ee;
node() {}
node(ll oo,ll oe,ll eo, ll ee):oo(oo),oe(oe),eo(eo),ee(ee) {}
}tree[N<<2];
node pushup(node a,node b)
{
node res;
res.oo = MAX(a.oo, b.oo, a.oo + b.eo, a.oe + b.oo);
res.oe = MAX(a.oe, b.oe, a.oo + b.ee, a.oe + b.oe);
res.eo = MAX(a.eo, b.eo, a.eo + b.eo, a.ee + b.oo);
res.ee = MAX(a.ee, b.ee, a.eo + b.ee, a.ee + b.oe);
return res;
}
void built(int i,int L,int R)
{
if(L == R)
{
if(L & 1)
scanf("%lld",&tree[i].oo), tree[i].oe = tree[i].eo = tree[i].ee = -inf;
else
scanf("%lld",&tree[i].ee), tree[i].oo = tree[i].oe = tree[i].eo = -inf;
return;
}
int mid = (L + R) >> 1;
built(lson,L,mid);
built(rson,mid+1,R);
tree[i] = pushup(tree[lson],tree[rson]);
}
void update(int i,int L,int R,int pos,ll val)
{
if(L == R)
{
if(L & 1)
tree[i].oo = val;
else
tree[i].ee = val;
return;
}
int mid = (L + R) >> 1;
if(pos <= mid)
update(lson,L,mid,pos,val);
else
update(rson,mid+1,R,pos,val);
tree[i] = pushup(tree[lson],tree[rson]);
}
node query(int i,int L,int R,int l,int r)
{
if(L >= l && R <= r)
return tree[i];
if(L > r || R < l)
return node(-inf,-inf,-inf,-inf);
int mid = (L + R) >> 1;
return pushup(query(lson,L,mid,l,r),query(rson,mid+1,R,l,r));
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d %d",&n,&m);
built(1,1,n);
while(m--)
{
int op,a,b;
scanf("%d %d %d",&op,&a,&b);
if(op)
update(1,1,n,a,b);
else
{
node ans = query(1,1,n,a,b);
printf("%lld\n",MAX(ans.oo,ans.oe,ans.eo,ans.ee));
}
}
}
return 0;
}
线段树专题
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