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POJ2155 Matrix

题解:

二维IUPQ题目

注意1点,不是直接在add时不是直接^=1,而是统计转换次数,这样在计算才能sum

代码:

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<map>#include<set>using namespace std;using namespace std;#define pb push_back#define mp make_pair#define se second#define fs first#define ll long long#define CLR(x) memset(x,0,sizeof x)#define SZ(x) ((int)(x).size())#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)typedef pair<int,int> P;const double eps=1e-9;const int maxn=100010;const int N=1000+10;const int mod=9901;ll read(){    ll x=0,f=1;char ch=getchar();    while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();}    while(ch>=0&&ch<=9){x=x*10+ch-0;ch=getchar();}    return x*f;}//-----------------------------------------------------------------------------int a[N][N],c[N][N];int n,m,x1,y1,x2,y2;char op;int lowbit(int x){return x&-x;}void add(int x,int y){    for(int i=x;i<=n;i+=lowbit(i))    for(int j=y;j<=n;j+=lowbit(j))    c[i][j]++;}int sum(int x,int y){    int cnt=0;    for(int i=x;i;i-=lowbit(i))    for(int j=y;j;j-=lowbit(j))    cnt+=c[i][j];    return cnt;}int main(){    int T;    scanf("%d",&T);    while(T--){        CLR(c);CLR(a);        scanf("%d%d",&n,&m);        for(int i=1;i<=m;i++){            scanf(" %c",&op);            if(op==C){                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                add(x1,y1);                add(x1,y2+1);                add(x2+1,y1);                add(x2+1,y2+1);            }            if(op==Q){                scanf("%d%d",&x1,&y1);                printf("%d\n",sum(x1,y1)%2);            }        }        if(T) printf("\n");    }    return 0;}

 

POJ2155 Matrix