题目链接:https://www.luogu.com.cn/problem/P2910
Floyd—Warshall算法的原理是动态规划。
设 (dis[i][j][k]) 为从 (i) 到 (j) 只以 (1 sim k) 中节点为中间结点的最短路径长度,则:
(1)若最短路径经过点 (k) ,那么 (dis[i][j][k]=dis[i][k][k-1]+dis[k][j][k-1])
(2)若最短路径不经过点 (k) ,那么 (dis[i][j][k]=dis[i][j][k-1])
因此 (dis[i][j][k]=min(dis[i][k][k-1]+dis[k][j][k-1], dis[i][j][k-1])) 。
如果我们把 (k) 放在最外层的循环,那么第三位在实现上可以省去。
实现代码如下:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 110, maxm = 10010;
int n, m, g[maxn][maxn], road[maxm], dist[maxn][maxn], ans;
const int INF = (1<<29);
int main() {
cin >> n >> m;
for (int i = 1; i <= m; i ++) cin >> road[i];
for (int i = 1; i <= n; i ++) {
for (int j = 1; j <= n; j ++) {
cin >> g[i][j];
dist[i][j] = (i == j) ? 0 : g[i][j];
}
}
for (int k = 1; k <= n; k ++)
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= n; j ++)
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
for (int i = 1; i < m; i ++)
ans += dist[ road[i] ][ road[i+1] ];
cout << ans << endl;
return 0;
}