poj 2029 get many persimmon trees

Posted 2020-10-09 jack_yyc

题目大意：

有一个01矩阵，求其所有a*b的子矩阵中总和最大的子矩阵的权值和

思路：

二维树状数组

nlogn裸题完事

虽然poj数据太弱了，二维前缀和就过了

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<queue>
 8 #include<vector>
 9 #include<set>
10 #include<stack>
11 #define inf 2147483611
12 #define ll long long
13 #define MAXN 510
14 using namespace std;
15 inline int read()
16 {
17     int x=0,f=1;
18     char ch;ch=getchar();
19     while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
20     while(isdigit(ch)) {x=x*10+ch-‘0‘;ch=getchar();}
21     return x*f;
22 }
23 int c[MAXN][MAXN],x,y,s,t,n,ans;
24 int lowbit(int x) {return x&(-x);}
25 void update(int a,int b)
26 {
27     for(int i=a;i<=y;i+=lowbit(i))
28         for(int j=b;j<=x;j+=lowbit(j))
29         {
30             c[i][j]++;
31         }
32 }
33 int sum(int a,int b)
34 {
35     int ans=0;
36     for(int i=a;i>=1;i-=lowbit(i))
37         for(int j=b;j>=1;j-=lowbit(j))
38         {
39             ans+=c[i][j];
40         }
41     return ans;
42 }
43 int main()
44 {
45     int a,b;
46     while(scanf("%d",&n)&&n)
47     {
48         memset(c,0,sizeof(c));
49         x=read(),y=read();
50         for(int i=1;i<=n;i++)
51         {
52             a=read(),b=read();
53             update(b,a);
54         }
55         s=read(),t=read();
56         ans=-1;
57         for(int i=t;i<=y;i++)
58             for(int j=s;j<=x;j++)
59             {
60                 ans=max(ans,sum(i,j)-sum(i-t,j)-sum(i,j-s)+sum(i-t,j-s));
61             } 
62         printf("%d\n",ans);
63     }
64 }

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