最短路径 POJ 1135 Domino Effect

题目链接：http://poj.org/problem?id=1135

代码参考：《图论算法理论、实现及应用》——北京大学出版社

题目大意：在多米诺骨牌中有几张关键牌，这些两张关键牌a，b之间有c张普通牌，问，推到第一张关键牌，多少时间可以把它全部推到，最后一张推到的牌，它的位置在哪里。

算法：dijkstra

在POJ上交了n遍G++WA，C++AC，伤不起

#include
#include
#include
using namespace std;

int n, m;
const int inf = 0x7ffffff;
int dist[1000], map[1000][1000];
int vis[1234];

void dijk(int k)
{
	int i, j;
	memset(vis, 0, sizeof(vis));
	for(i = 1; i <= n; i ++)
	{
		dist[i] = map[k][i];
	}
	dist[k] = 0;
	vis[k] = 1;
	for(i = 1; i < n; i ++)
	{
		int minn = inf;
		for(j= 1; j <= n; j ++)
		{
			if(!vis[j] && dist[j] < minn)
			{
				minn = dist[j];
				k = j;
			}
		}
		if(minn == inf)
			break;
		vis[k] = 1;
		for(j = 1; j <= n; j ++)
		{
			if(!vis[j] && map[k][j] < inf && dist[j] > map[k][j] + dist[k])
				dist[j] = map[k][j] + dist[k];
		}
	}
	double maxn = -inf ;
	int pos;
	for(i = 1; i <= n; i ++)
		if(maxn < dist[i])
		{
			pos = i;
			maxn = dist[i];
		}
	double maxm = -inf;
	int pos1, pos2;
	for(i = 1; i < n; i ++)
		for(j = i+1; j <= n; j ++)
			if(map[i][j] < inf && (dist[i] + dist[j] + map[i][j]) / 2.0 > maxm)
			{
				pos1 = i, pos2 = j;
				maxm = (dist[i] + dist[j] + map[i][j]) / 2.0;
			}
	if(maxn >= maxm )
	{
		printf("%.1lf seconds, at key domino %d.\n\n", maxn, pos);
	}
	else
		printf("%.1lf seconds, between key dominoes %d and %d.\n\n", maxm, pos1, pos2);
//			printf("pos2 = %d\n", pos2);
}


int main()
{
	int ica = 1;
	while(scanf("%d%d", &n, &m))
	{
		if(!n && !m)
			break;
		int i, j;
		for(i= 1; i <= n; i ++)
		for(j = 1; j <= n; j ++)
		{
			map[i][j] = inf;
		}
		int a, b, c;
		for(i = 1; i <= m ; i ++)
		{
			scanf("%d%d%d", &a, &b, &c);
			map[a][b] = c;
			map[b][a] = c; 
		}
		printf("System #%d\nThe last domino falls after ", ica ++ );
		dijk(1);
	}
	return 0;
}