ZOJ 1952 Heavy Cargo （dijkstra )

又一次深刻的理解了dijkstra，他的各种变形可以解决各种各样的问题，有时候我们用dij不仅可以求出最小值，也可以求出符合要求的最大值。只要之后更新的东西不会改变已经加入到集合S中的点 。

 

#include<stdio.h> #include<math.h> #include<string.h> #define INF 99999999 int count = 0; char name[201][40]; int find(char *a) { 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( void) { int n,r,t,max,l=1; int start,end,j,i,now; int from,to; int mat[201][201]; int dist[201],flag[201]; char a[40],b[40]; while( scanf("%d%d",&n,&r) != EOF ) { if( !n && !r ) break; for( i = 1; i <= n; i++ ) { dist[i] = 0; flag[i] = 0; for( j = 1; j <= n; j++ ) mat[i][j] = 0; } for( i = 1; i <= r; i++ ) { scanf("%s%s",a,b); from = find(a); to = find(b); scanf("%d",&j); mat[from][to] = mat[to][from] = j; } scanf("%s%s",a,b); start = find(a); end = find(b); dist[now = start] = INF; flag[now] = 1; for(i = 1; i < n; i++ ) { for( j = 1; j <= n; j++ ) { if( flag[j] ) continue; t = dist[now]<mat[now][j]?dist[now]:mat[now][j]; if( dist[j] < t ) dist[j] = t; } for( j = 1,max = -1; j <= n; j++ ) if( flag[j] == 0 && dist[j] > max ) max = dist[now=j]; flag[now] = 1; } printf("Scenario #%d/n",l++); printf("%d tons/n/n",dist[end]); } return 0; }