1. 时间复杂度O(n^3)
2. 动态规划
3. d[k][i][j] = min(d[k-1][i][j],d[k-1][i][k] + d[k-1][k][j])
4. 可以简化为
5. d[i][j] = min(d[i][j], d[i][k] + d[j][k]) //考虑经过k点时的最短路

代码

#include<iostream>
#include<algorithm>
#include<cstdio>using namespace std;const int N = 210;int g[N][N];int n;void floyed(){for(int k = 1; k <= n; k++){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){g[i][j] = min(g[i][j], g[i][k] + g[k][j]);}}}}
int main(){int m,k;cin>>n>>m>>k;int x, y, z;//考虑重边和自环for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(i == j){g[i][j] = 0; //去掉环 }else g[i][j] = 0x3f3f3f3f;}}for(int i = 0; i < m; i++){cin>>x>>y>>z;g[x][y] = min(g[x][y],z);}int a, b;int ans;floyed();for(int i = 0; i < k; i++){cin>>a>>b;if(g[a][b] > 0x3f3f3f3f/2){cout<<"impossible" << '\n';}else cout<< g[a][b]<< '\n';}return 0;
}