题解 | #矩阵的最小路径和#

#include<iostream>
using namespace std;
int graph[501][501],dp[501][501];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,m;
    cin >> n >> m;
    for(int i = 1;i <= n;++i)
        for(int j = 1;j <= m;++j){
            cin >> graph[i][j];
            if(i == 1 && j == 1)
                dp[i][j] = graph[i][j];
            else if(i == 1)
                dp[i][j] = dp[i][j - 1] + graph[i][j];
            else if(i >= 2 && j != 1)
                dp[i][j] = min(dp[i - 1][j] + graph[i][j],dp[i][j - 1] + graph[i][j]);
            else 
                dp[i][j] = dp[i - 1][j] + graph[i][j];
        }
    cout << dp[n][m];
}