ZOJ Problem Set - 1092 Arbitrage (Floyd)

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Floyd-Warshall算法描述：

For k←1 to n do // k为“媒介节点”     For i←1 to n do        For j←1 to n do           if (dist(i,k) + dist(k,j) < dist(i,j)) then // 是否是更短的路径？              dist(i,j) = dist(i,k) + dist(k,j)

由于本题是求最大的收益，因此描述如下：

for (k = 1; k <= n; k++)for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)if (map[i][j] < map[i][k] * map[k][j])map[i][j] = map[i][k] * map[k][j];

解题如下：

#include <stdio.h>#include <stdlib.h>#include <string.h>double map[32][32];int n;void input(int N){int j, i, x, y, m, flag;double price;char name[32][100];char temp1[100], temp2[100];for (i = 1; i <= N; i++)scanf("%s", name[i]);scanf("%d", &m);for (i = 1; i <= N; i++){for (j = 1; j <= N; j++)map[i][j] = 0.0;map[i][i] = 1.0;}while (m--){scanf("%s %lf %s", temp1, &price, temp2);flag = 0;for (i = 1; i <= N && flag != 2; i++){if (strcmp(temp1, name[i]) == 0){x = i;flag++;}if (strcmp(temp2, name[i]) == 0){y = i;flag++;}}map[x][y] = price;}}void floyd(int *success){int k, i, j;for (k = 1; k <= n; k++)for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)if (map[i][j] < map[i][k] * map[k][j])map[i][j] = map[i][k] * map[k][j];for (k = 1; k <= n; k++)if (map[k][k] > 1){*success = 1;break;}}int main(){int k, i, j;int success, count = 1;while (scanf("%d", &n) && n){success = 0;input(n);floyd(&success);if (!success)printf("Case %d: No\n", count++);elseprintf("Case %d: Yes\n", count++);}return 0;}

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