P2910 [USACO08OPEN]Clear And Present Danger S题解

bfac3402-1150-46c7-8798-f545389dbac3
其使用最简单的方法做就好了忘了啥名字，类似于dp
给个代码

#include<bits/stdc++.h>
using namespace std;
long long n,m;
long long c[105][105];
long long ans;
short l[100000];
int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) cin >> l[i];
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> c[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			for (int k = 1; k <= n; k++) {
				c[j][k] = min(c[j][k],c[j][i] + c[i][k]);
			}
		}
	}
	for(int i = 2; i <= m; i++) {
		ans += c[l[i - 1]][l[i]];
	}
	cout << ans << endl;
	return 0;
}
/*
3 4
1
2
1
3
0 5 1
5 0 2
1 2 0 

7
*/

注意：是for这样的

~~~~~~~~~~~~~

【重要】

------中转点
----------第一个点
--------------第二个点
再给个模板

for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			for (int k = 1; k <= n; k++) {
				c[j][k] = min(c[j][k],c[j][i] + c[i][k]);
			}
		}
	}