floyed-多源汇聚最短路

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
#include
#include

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;
}