Floyd-Warshall算法及其应用

1.递归方程

d_ij(k)=w_ij(k = 0), min(d_ij(k-1), d_ik(k-1)+d_kj(k-1)(k>=1)

2.程序

 /**************************************Algorithm: Floyd (With Direction)*************************************/#include <iostream>using namespace std;const int MAXV = 100;//The number of vertexsconst int MAX = 1 << 28;//Don't make it too big as we use additionint map[MAXV][MAXV], dis[MAXV][MAXV], pi[MAXV][MAXV];//map record the direct distance, dis record the minest distanceint len, ans[MAXV];int n, m, u, v, w;//n denotes the number of vertexs, m denotes the number of edgesint main() {scanf("%d%d", &n, &m);//INITIALIZEfor (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (i == j) map[i][j] = 0; else map[i][j] = MAX;//READ THE EDGEfor (int i = 0; i < m; i++) {scanf("%d%d%d", &u, &v, &w);if (w < map[u][v]) map[u][v] = w;//map[u][v] = map[v][u] = w;}//INITIALIZEfor (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {dis[i][j] = map[i][j];if (dis[i][j] != MAX) pi[i][j] = i; else pi[i][j] = 0;}//FLOYED-WARSHALLfor (int k = 1; k <= n; k++)for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (dis[i][k]+dis[k][j] < dis[i][j]) {dis[i][j] = dis[i][k] + dis[k][j];pi[i][j] = pi[k][j];}//PRINT THE SHORTEST DISTANCES printf("Please input the vertexs to show the SP:");scanf("%d %d", &u, &v);len = 0;while (u != v) {ans[len++] = v;v = pi[u][v];}ans[len++] = u;for (int i = len-1; i >= 0; i--) printf("%d/n", ans[i]);//NOTE that print in the reverse orderreturn 0;}

3.应用：

1.求最小环USACO4.1.3

具体的程序见USACO专题，基本的代码如下：

/**************************************************************Algorithm: Floyd (With direction) to get the minimal cycle*************************************************************/#include <iostream>using namespace std;const int MAXV = 100;//The number of vertexsconst int MAX = 1 << 28;//Don't make it too big as we use additionint map[MAXV][MAXV], dis[MAXV][MAXV], pi[MAXV][MAXV];//map record the direct distance, dis record the minest distanceint min_cycle, len, ans[MAXV];int n, m, u, v, w;//n denotes the number of vertexs, m denotes the number of edgesvoid recordpath(int i, int j, int k) {//i -> j -> klen = 0;while (j != i) {ans[len++] = j;j = pi[i][j];}ans[len++] = i; ans[len++] = k;}int main() {scanf("%d%d", &n, &m);//INITIALIZEfor (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (i == j) map[i][j] = 0; else map[i][j] = MAX;//READ THE EDGEfor (int i = 0; i < m; i++) {scanf("%d%d%d", &u, &v, &w);if (w < map[u][v]) map[u][v] = w;//map[u][v] = map[v][u] = w;}//INITIALIZEfor (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++) {dis[i][j] = map[i][j];if (dis[i][j] != MAX) pi[i][j] = i; else pi[i][j] = 0;}//FLOYED-WARSHALLmin_cycle = MAX;for (int k = 1; k <= n; k++) {for (int i = 1; i < k; i++)if (map[k][i] < MAX)for (int j = 1; j < k; j++)if (map[j][k] < MAX && dis[i][j] + map[j][k] + map[k][i] < min_cycle) {min_cycle = dis[i][j] + map[j][k] + map[k][i];recordpath(i, j, k);}for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (dis[i][k]+dis[k][j] < dis[i][j]) {dis[i][j] = dis[i][k] + dis[k][j];pi[i][j] = pi[k][j];}}//PRINT THE SHORTEST DISTANCES printf("The minimal cycle: %d/nThe vertex on the cycle: %d/n", min_cycle, len);for (int i = len-1; i >= 0; i--) printf("%d/n", ans[i]);return 0;}

2.URAL 1004