PAT Emergency dijsktra

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/sinat_22659021/article/details/52421799

首先wa了好多次，只有第一个答案是正确的，后来才知道原来是第一个输出是最短路径的个数，一直以为是最短路径。。。。无语了，代码中d[i]代表从c1到节点i的最短路径个数，sum[i]代表最短路径中能集结的最多救援队数，在每次更新节点路径长度的时候要考虑对他们该如何操作。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define Max 0x7fffffff - 1000000
long long int gra[505][505];
int num[505];
int d[505];
int n,m,c1,c2,counter;
int visit[505];
int sum[505];

int dijsktra()
{
	int i,j;
	memset(visit,0,sizeof(visit));
	visit[c1] = 1;
	int now = c1;
	int min;
	for(i=0; i<n; i++)
		sum[i] = num[i];
	sum[c1] = num[c1];
	for(j=0; j<n; j++)
	{
		for(i=0; i<n; i++)
		{
			if(!visit[i]&& gra[c1][i] >= gra[c1][now]+gra[now][i])
			{
				if(gra[c1][i] == gra[c1][now]+gra[now][i])
				{
					if(now == c1)
					{
						if(gra[c1][i] == Max)
							d[i] = 0;
						else
							d[i] = 1;
					}	
					else
						d[i]+= d[now];
					if(sum[now]+num[i] > sum[i])
						sum[i] = sum[now] + num[i];
				}
				else
					{
						sum[i] = sum[now]+num[i];
						gra[c1][i] = gra[c1][now]+gra[now][i]; 	
						d[i] = d[now];
					}
			}
		}
		
		min = Max;
		for(i=0; i<n; i++)
		{
			if(!visit[i] && gra[c1][i] < min)
			{
				min = gra[c1][i];
				now = i;
			}
		}
		visit[now] = 1;
		if(now == c2)
			break;
	}
	return 0;
}
int main()
{
	int i,j,a,b,c;
	scanf("%d%d%d%d",&n,&m,&c1,&c2);
	for(i=0; i<n; i++)
		scanf("%d",&num[i]);	
	for(i=0; i<n; i++)
		for(j=0; j<n; j++)
		{
			gra[i][j] = Max;
			if(i == j)
				gra[i][j] = 0;
		} 		
	for(i=0; i<m; i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		gra[a][b] = gra[b][a] = c;
	}
	if(c1 == c2)
	{
		printf("%d %d\n",1,num[c2]);
		return 0;
	}	
	dijsktra();
	printf("%d %d\n",d[c2],sum[c2]);
	return 0; 
}