題目：
http://acm.swust.edu.cn/#/problem/1132/490
題目描述
Several coins are placed in cells of an n×m board. A robot, located in the upper left cell of the board, needs to collect as many of the coins as possible and bring them to the bottom right cell. On each step, the robot can move either one cell to the right or one cell down from its current location.
輸入
The fist line is n,m, which 1< = n,m <= 1000.
Then, have n row and m col, which has a coin in cell, the cell number is 1, otherwise is 0.
輸出
The max number Coin-collecting by robot.
樣例輸入
5 6
0 0 0 0 1 0
0 1 0 1 0 0
0 0 0 1 0 1
0 0 1 0 0 1
1 0 0 0 1 0
樣例輸出
5
說明：
簡單dp，從左上到右下逐步轉移狀態即可。
代碼：
#include<bits/stdc++.h>
using namespace std;
const int maxn=1005;
int n,m;
int a[maxn][maxn],dp[maxn][maxn];
int main(){
cin>>n>>m;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>a[i][j];
}
}
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
dp[i][j]=max(dp[i-1][j],dp[i][j-1])+a[i][j];
}
}
cout<<dp[n][m]<<endl;
return 0;
}