ZOJ 1952( Dijkstra )要求卡车的最大载货量，即是求dist[]的最小值这里关键是把字符串转化为数字存储在邻接矩阵cost[][]中.开始看了党姐的代码不懂，又看了一遍，大悟！

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 90000000

int count = 0;  
char name[201][40];  
int find(char *a)  //此函数实现将字符串转化为数字，剽窃党姐的呵呵>_<
{  
    int i;  
    for( i = 1; i <= count; i++ )  
        if( strcmp(a,name[i]) == 0 )  
            return i;  
    count = i;  
    strcpy(name[i],a);  
    return i;  
}  

int main()
{
    int n, r, l, s[201], dist[201], cost[201][201];
    char a[40], b[40];
    int i, j, start, end, now, k=1;
    int temp, max, from, to;
    
    while(scanf("%d%d",&n,&r) != EOF)
    {
        if (n==0 && r==0) break;
        for(i=1; i<=n; i++)
        {
           dist[i] = 0;
           s[i] = 0;
           for(j=1; j<=n; j++)
           {
              cost[i][j] = 0;      
           }
        }
        for(i=1; i<=r; i++)
        {
           scanf("%s%s",a,b);
           from = find(a);      
           to = find(b);
           scanf("%d",&l);
           cost[from][to] = cost[to][from] = l;
        }
        scanf("%s%s",a,b);
        start = find(a);
        end = find(b);
                
        now = start;         
        dist[now] = MAX;     
        s[now] = 1;
        for(i=1; i<n; i++)
        {      
           max = -1;    
           for(j=1; j<=n; j++)
           {
             if (s[j] == 0)
               temp = dist[now]<cost[now][j] ? dist[now]:cost[now][j];
             if(dist[j] < temp)
               dist[j] = temp; 
           }
           for(j=1; j<=n; j++)
             if(s[j]==0 && dist[j] > max)
             {
                now = j;
                max = dist[now];
             }
           s[now] = 1;     
        }
        printf("Scenario #%d\n",k++);
        printf("%d tons\n\n",dist[end]);
    }   
    return 0;
}