题解 | #矩阵的最小路径和#
矩阵的最小路径和
http://www.nowcoder.com/practice/38ae72379d42471db1c537914b06d48e
#include<bits/stdc++.h>
using namespace std;
const int maxn = 520;
int n, m, mp[maxn][maxn], f[maxn][maxn];
int main(){
// freopen("11_minsum.in", "r", stdin);
cin >> n >> m;
memset(f, 0x3f, sizeof(f)); // 防止选到边界元素(如 i == 0 || j == 0 上的元素
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++){
cin >> mp[i][j];
if (i == 1 && j == 1) f[i][j] = mp[1][1]; // 边界条件
// f[i][j] = 左边 + 自己 || 上边 + 自己,取较小值,不断往右下角计算
else f[i][j] = min(f[i - 1][j], f[i][j - 1]) + mp[i][j];
}
}
cout << f[n][m] << endl;
return 0;
}