POJ2029 Get Many Persimmon Trees

  题目链接：http://poj.org/problem?id=2029

  二维树状数组+容斥

  c[i][j]的意义是(i，j)为右下角的前缀和

  注意要向上统计，向下修改

  注意区别单点修改与区间修改

  贴代码

#include<cstdio>#include<cstring>using namespace std;int const maxn=105;int c[maxn][maxn];int n,w,h;int lowbit(int x){return x&-x;}void add(int x,int y){for (int i=x;i<=w;i+=lowbit(i))for (int j=y;j<=h;j+=lowbit(j))c[i][j]+=1;}int query(int x,int y){int sum=0;for (int i=x;i>=1;i-=lowbit(i))for (int j=y;j>=1;j-=lowbit(j))sum+=c[i][j];return sum;}int max(int x,int y){return x>y?x:y;}int main(){//freopen("2029.in","r",stdin);//freopen("2029.out","w",stdout);scanf("%d",&n);while (n){scanf("%d%d",&w,&h);w++;h++;memset(c,0,sizeof(c));while (n--){int x,y;scanf("%d%d",&x,&y);x++;y++;add(x,y);}int s,t;scanf("%d%d",&s,&t);int ans=0;for (int i=s+1;i<=w;i++)for (int j=t+1;j<=h;j++) ans=max(ans,query(i,j)-query(i-s,j)-query(i,j-t)+query(i-s,j-t));printf("%d\n",ans);scanf("%d",&n);}return 0;}

【写的有漏洞的，欢迎路过大神吐槽】

  2017/07/06 22:03:04

  Ending.