最短路之Floyd算法

/* Floyd算法 */

/************************************************************************
   Floyd算法的核心可用以下代码表示

    for (int k = 0; k < n; ++k){
        for (int i = 0; i < n; ++i){
            for (int j = 0; j < n; ++j){
                if (d[i][j] < INF && d[k][j] < INF){
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
                }//if
            }//for(j)
        }//for(i)
    }//for(k)
    时间复杂度: O(n^3)
 ************************************************************************/

#include <cstdio>

int n, m;
const int maxn = 100;
const int INF = 65535;
int w[maxn][maxn];
int d[maxn][maxn];
int p[maxn][maxn];

void Floyd(){
    for (int i = 0; i < n; ++i){
        for (int j = 0; j < n; ++j){
            d[i][j] = w[i][j];
            p[i][j] = j;
        }//for(j)
    }//for(i)

    for (int k = 0; k < n; ++k){
        for (int i = 0; i < n; ++i){
            for (int j = 0; j < n; ++j){
                if (d[i][j] > d[i][k] + d[k][j]){
                    d[i][j] = d[i][k] + d[k][j];
                    p[i][j] = p[i][k];
                }
            }//for(j)
        }//for(i)
    }//for(k)
}

int main()
{
    int from, to, v;
    scanf("%d%d", &n, &m);
    //初始化权值
    for (int i = 0; i < n; ++i){
        for (int j = 0; j < n; ++j){
            w[i][j] = INF;
        }
    }
    for (int i = 0; i < n; ++i){
        w[i][i] = 0;
    }
    //边的输入
    for (int i = 0; i < m; ++i){
        scanf("%d%d%d", &from, &to, &v);
        w[from][to] = w[to][from] = v;
    }
    Floyd();

    for (int i = 0; i < n; ++i){
        for (int j = 0; j < n; ++j){
            printf("%3d", d[i][j]);
        }
        printf("\n");
    }

    return 0;
}