题目描述
输入
输出
样例输入
3 2
1 9 8
3 2 0
1 8 3
9 8 4
0 5 15
1 9 6
1 1 3 3
2 3 1 2
样例输出
661
388
题解
前缀和
不妨设a<=c，b<=d，那么$\ \ \ \sum\limits_{i=a}^c\sum\limits_{j=b}^dR[i][j]\\=\sum\limits_{i=a}^c\sum\limits_{j=b}^d\sum\limits_{k=1}^nP[i][k]·Q[k][j]\\=\sum\limits_{k=1}^n(\sum\limits_{i=a}^cP[i][k])·(\sum\limits_{j=b}^dQ[k][j])\\=\sum\limits_{k=1}^n(sumP[c][k]-sumP[a-1][k])(sumQ[k][d]-sumQ[k][b-1])$
时间复杂度$O(nm)$，可过。
#include <cstdio>
#include <algorithm>
using namespace std;
#define N 2010
int p[N][N] , q[N][N] , sp[N][N] , sq[N][N];
inline int read()
{
int ret = 0; char ch = getchar();
while(ch < ‘0‘ || ch > ‘9‘) ch = getchar();
while(ch >= ‘0‘ && ch <= ‘9‘) ret = (ret << 3) + (ret << 1) + ch - ‘0‘ , ch = getchar();
return ret;
}
int main()
{
int n , m , i , j , a , b , c , d;
long long ans;
n = read() , m = read();
for(i = 1 ; i <= n ; i ++ )
for(j = 1 ; j <= n ; j ++ )
a = read() , sp[i][j] = sp[i - 1][j] + a;
for(i = 1 ; i <= n ; i ++ )
for(j = 1 ; j <= n ; j ++ )
a = read() , sq[i][j] = sq[i][j - 1] + a;
while(m -- )
{
a = read() , b = read() , c = read() , d = read();
if(a > c) swap(a , c);
if(b > d) swap(b , d);
ans = 0;
for(i = 1 ; i <= n ; i ++ ) ans += (long long)(sp[c][i] - sp[a - 1][i]) * (sq[i][d] - sq[i][b - 1]);
printf("%lld\n" , ans);
}
return 0;
}
原文：http://www.cnblogs.com/GXZlegend/p/7110298.html