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HDU 3954 Level up 线段树

2024-07-18 20:20:00   227人阅读

～～～NotOnlySuccess 出的题～～～

看了题之后觉得和HDU 4027有点像，给的K很小，只有10，目测只要有人升级的时候直接更新到叶子节点就ok了。不过不同于HDU 4027 的是，那题每一次更新都相当于这题的一次升级操作，这题里面可能会出现一次操作之后没有升级和出现升级两种情况，一时半会没了思路。

无奈去搜题解，发现我只要维护一个区间当中距离升级最近的人所需要的基础升级经验，即不算等级加成的裸的升级经验，如果在一次涨经验之后，出现当前区间当中有人会升级，直接将每一个要升级的人更新到叶子节点即可。 复杂度的话，每个人最多升级K次，应该撑死也就O（NKlogN）的复杂度吧，可以接受。

#include <cstdio>#include <cstring>#include <iostream>#include <map>#include <set>#include <vector>#include <string>#include <queue>#include <deque>#include <bitset>#include <list>#include <cstdlib>#include <climits>#include <cmath>#include <ctime>#include <algorithm>#include <stack>#include <sstream>#include <numeric>#include <fstream>#include <functional>using namespace std;#define MP make_pair#define PB push_back#define lson rt << 1, l, mid#define rson rt << 1 | 1, mid + 1, rtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int,int> pii;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 1e4 + 50;const int maxk = 15;int lazy[maxn << 2], min_need[maxn << 2], max_exp[maxn << 2], max_level[maxn << 2];int need_exp[maxk], N, M, K;int count_exp(int exp,int level) {    int tar = need_exp[level + 1],ret = (tar - exp) / level;    if((tar - exp) % level != 0) ret++;    return ret;}void pushup(int rt) {    int lc = rt << 1, rc = rt << 1 | 1;    max_level[rt] = max(max_level[lc], max_level[rc]);    max_exp[rt] = max(max_exp[lc], max_exp[rc]);    min_need[rt] = min(min_need[lc], min_need[rc]);}void pushdown(int rt) {    if(lazy[rt] == 0) return;    int lc = rt << 1, rc = rt << 1 | 1;    lazy[lc] += lazy[rt]; lazy[rc] += lazy[rt];    min_need[lc] -= lazy[rt]; min_need[rc] -= lazy[rt];    max_exp[lc] += lazy[rt] * max_level[lc];    max_exp[rc] += lazy[rt] * max_level[rc];    lazy[rt] = 0;}void build(int rt,int l,int r) {    lazy[rt] = 0;    if(l == r) {        max_exp[rt] = 0;        max_level[rt] = 1;        min_need[rt] = need_exp[2];    }    else {        int mid = (l + r) >> 1;        build(lson); build(rson);        pushup(rt);    }}//升级操作void level_up(int rt) {    while(max_exp[rt] >= need_exp[max_level[rt] + 1]) max_level[rt]++;    min_need[rt] = count_exp(max_exp[rt],max_level[rt]);}void update(int rt,int l,int r,int ql,int qr,int Val) {    int mid = (l + r) >> 1;    if(ql <= l && qr >= r) {        lazy[rt] += Val; min_need[rt] -= Val;        max_exp[rt] += max_level[rt] * Val;        if(min_need[rt] <= 0) {            if(l == r) level_up(rt);            else {                pushdown(rt);                update(lson,ql,qr,0);                update(rson,ql,qr,0);                pushup(rt);            }        }    }    else {        pushdown(rt);        if(ql <= mid) update(lson,ql,qr,Val);        if(qr > mid) update(rson,ql,qr,Val);        pushup(rt);    }}int query(int rt,int l,int r,int ql,int qr) {    if(ql <= l && qr >= r) return max_exp[rt];    int mid = (l + r) >> 1, ret = -1;    pushdown(rt);    if(ql <= mid) ret = max(ret, query(lson, ql, qr));    if(qr > mid) ret = max(ret, query(rson, ql, qr));    return ret;}int main() {    int T; scanf("%d", &T);    for(int kase = 1; kase <= T; kase++) {        char cmd[3];        int l,r,val;        printf("Case %d:\n",kase);        scanf("%d%d%d",&N,&K,&M);        for(int i = 2;i <= K;i++) scanf("%d",&need_exp[i]);        need_exp[K + 1] = INT_MAX;        build(1,1,N);        for(int i = 1;i <= M;i++) {            scanf("%s",cmd);            if(cmd[0] == ‘W‘) {                scanf("%d%d%d", &l, &r, &val);                update(1, 1, N, l, r, val);            }            else {                scanf("%d%d",&l,&r);                printf("%d\n",query(1, 1, N, l, r));            }        }        puts("");    }    return 0;}

　　


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