UVALive 7079 - How Many Maos Does the Guanxi Worth(最短路Floyd)
题目大意:n个人编号从1到n,m条关系,a找b帮忙或b找a帮忙需要花费c元,当然a可以通过d找b帮忙(即a找d帮忙,d再找b帮忙)
现在1号要找n号帮忙,你需要去找1号和n号之外的人,让他不帮忙,即破坏这条关系链,如果1号最终能找n号b帮忙则输出过程中所
用的最大花费,否则输出Inf
分析:
转化为最短路问题:a到b的花费当做a点到b点的距离
破坏关系链只需在2—n-1号这n-3点中依次找每一个点,将该点到其他点和其他点到该点的距离令为INF,然后找一个1到n的最小距离,
得到n-3个距离,输出这n-3个距离中最大的距离
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm> using namespace std; const int N = ;
const int INF = 0x3f3f3f3f; int G[N][N], maps[N][N];
int n; void Init()
{
int i, j;
for(i = ; i <= n ; i++)
{
for(j = ; j <= n ; j++)
{
if(i == j)
G[i][j] = ;
else
G[i][j] = G[j][i] = INF;
}
}
} int Floyd(int m)
{
int i, j, k;
for(i = ; i <= n ; i++)
{
for(j = ; j <= n; j++)
{
if(i == m || j == m)
maps[i][j] = maps[j][i] = INF;//让编号为m的人不帮,来破坏该条关系链
else
maps[i][j] = maps[j][i] = G[i][j];
}
}
for(k = ; k <= n ; k++)
{
for(i = ; i <= n ; i++)
{
for(j = ; j <= n ; j++)
{
if(maps[i][j] > maps[i][k] + maps[k][j])
maps[i][j] = maps[i][k] + maps[k][j];
}
}
}
return maps[][n];
} int main()
{
int m, i, a, b, c;
while(scanf("%d%d", &n, &m), m + n)
{
Init();
while(m--)
{
scanf("%d%d%d", &a, &b, &c);
G[b][a] = G[a][b] = c;
}
int ans = ;
for(i = ; i <= n - ; i++)
ans = max(ans, Floyd(i));
if(ans == INF)
printf("Inf\n");
else
printf("%d\n", ans);
}
return ;
}
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