UVA 567 By ACReaper

这道题目，坑死我了，特别是数据输出的时候有句话 The test result line should contain the start country code A right-justified in columns 1 and 2; the string `` to " in columns 3 to 6; 呜呜，害哥格式错误，找了个半天，你妈呀！！！。题目的大意思就是给你一个图，求任意两点之间，要连同所要征服的国家的最少数目，其实这就是最短路问题。由于是任意两点之间，当然采用Floyd-Warshall算法。呜呜。复习下其伪代码：

d[i][j] = 0 (i == j)

d[i][j] = d[i][k] + d[k][j] (i != j)

Floyd-Warshall（）｛

    n = rows(W);

    D(0) = W;

   for(int k = 1; k <= n; k++){

for(int i = 1; i <=n ;i++){

for(int j = 1; j <= n; j++)

d[i][j] <?= d[i][k] + d[k][j];

}

   }

｝

#include <stdio.h>
#define MAXN 50
#define INF 1 << 25
//#define TEST
#define OJ
int w[MAXN][MAXN];
int d[MAXN][MAXN];
int main(){
	#ifdef TEST
	freopen("X:in.txt","r",stdin);
	freopen("X:out567.txt","w",stdout);
	#endif
	int n;
	int testcase = 1;
	while(1){
		bool flag = false;
		for(int i = 1; i <= 20; i++){
			for(int j = 1; j <= 20; j++){
				w[i][j] = d[i][j] = i == j?0:INF;
			}
		}
		bool first = true;
		for(int j = 1; j <= 19; j++){//结点V1---V19 
			int m,v;
			if(scanf("%d",&m)== EOF){
				flag = true;
				break;
			}
			for(int i = 1; i <= m; i++){//每个结点按顺序的邻接点 
				scanf("%d",&v);
				d[j][v] = d[v][j] = w[j][v] =w[v][j] = 1;
			}
		}	
		if(flag)
	 		break;
		else{
			//Graph-algo-count;
			for(int k = 1; k <= 20; k++){
				for(int i = 1; i <= 20; i++){
					for(int j = 1; j <= 20; j++){
						d[i][j] = d[i][j] < d[i][k] + d[k][j] ? d[i][j] : d[i][k] + d[k][j] ;
					}
				}
			}
			printf("Test Set #%d\n",testcase);
			scanf("%d",&n);
			for(int i = 1; i <= n ; i++){
				int u,v,temp;
				scanf("%d%d",&u,&v);
				printf("%2d to %2d: %d\n",u,v,d[temp = u > v?v:u][u + v - temp]);
			}
			printf("\n");
			testcase++;
		}
	}
	return 0;
}