poj 2263 Floyd

一开始又是试了一下求最长路径+路径记录，之后又试了一下权值取反求最短路径+记录……发现都行不通= =，之后bing了一下发现Floyd貌似不能出路有环的最长路径…………后来看书发现，原来直接改一下递推公式就变成了求“最大容量路”，即此题的算法！

灵活，灵活啊~

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
const int POINT = 210;
int A[POINT][POINT];
string city[POINT];
int main()
{
    int cases = 0;
    int n,r;
    while(scanf("%d %d",&n,&r),n||r)
    {
        cases++;
        memset(A,0,sizeof(A));
        int numcity = 1;
        for(int i=1; i<=r; i++)
        {
            string a,b;
            int w;
            cin>>a>>b>>w;
            int u=0,v=0;
            for(int j=1; j<=numcity; j++)
            {
                if(a==city[j])
                    u = j;
                if(b==city[j])
                    v = j;
            }
            if(u==0)
            {
                u = numcity;
                city[numcity++] = a;
            }
            if(v==0)
            {
                v = numcity;
                city[numcity++] = b;
            }
            A[u][v] = A[v][u]  = w;
        }
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                for(int k=1; k<=n; k++)
                        A[j][k] = max(A[j][k],min(A[j][i],A[i][k]));
        string a,b;
        cin>>a>>b;
        int u,v;
        for(int i=1; i<=n; i++)
        {
            if(a == city[i])
                u = i;
            if(b == city[i])
                v = i;
        }
        printf("Scenario #%d\n%d tons\n\n",cases,A[u][v]);
        for(int i=1;i<=n;i++)
            city[i].clear();
    }
}