题意：给出n*m网格中每个格子有A，B两种矿，A矿必须从右向左运输，B矿必须从下往上运输，管子不能拐弯或者折断，要求收集到的A，B矿的总量尽量大
思路：对于每个格子不是安A就是B矿的管道，所以只有两个选择。
所以dp[i][j]=max(dp[i][j-1]+b[i][j],dp[i-1][j]+a[i][j])，其中a[i][j],b[i][j]分别代表从右往左和从下往上运输方式的前j,前i的总矿数
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 505;
int a[MAXN][MAXN],b[MAXN][MAXN],dp[MAXN][MAXN];
int n,m;
int main(){
while (scanf("%d%d",&n,&m) != EOF && n+m){
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++){
scanf("%d",&a[i][j]);
a[i][j] += a[i][j-1];
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++){
scanf("%d",&b[i][j]);
b[i][j] += b[i-1][j];
}
memset(dp,0,sizeof(dp));
int ans = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
dp[i][j] = max(dp[i-1][j]+a[i][j],dp[i][j-1]+b[i][j]);
printf("%d\n",dp[n][m]);
}
return 0;
}