ZOJ Problem Set - 1092 Arbitrage (Floyd)

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)

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];

#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++);
else
printf("Case %d: Yes\n", count++);
}
return 0;
}