nyoj 247 虚拟的城市之旅（两次SPFA求两点最大差值）

思路：
1>   求 1 到所有点的最短路径， 
2>   求 n 到所有点的最长路径
3>   做差。 

看是否可达，判断是否入过队。 

#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAXN 100004
#define INF 0x3f3f3f3f

int dist1[MAXN],dist2[MAXN];
int vis1[MAXN],vis2[MAXN];
int head1[MAXN],head2[MAXN];
int top;

struct Edge{
	int to,next;
}edge[MAXN*5];

void addedge(int u,int v){
	edge[top].to=v;
	edge[top].next=head1[u];
	head1[u]=top++;
	
	edge[top].to=u;
	edge[top].next=head2[v];
	head2[v]=top++;
}
void SPFA1(int st){
	memset(vis1,0,sizeof(vis1));
	vis1[st]=1;
	queue<int>q;
	q.push(st);
	
	while(!q.empty()){
		int u = q.front();
		q.pop();
		for(int i=head1[u];i!=-1;i=edge[i].next){
			int v=edge[i].to;
			dist1[v]=min(dist1[v],dist1[u]);
			if(!vis1[v]){
				vis1[v]=1;
				q.push(v);
			}	
		}
	}
}
void SPFA2(int st){
	memset(vis2,0,sizeof(vis2));
	vis2[st]=1;
	queue<int>q;
	q.push(st);
	
	while(!q.empty()){
		int u = q.front();
		q.pop();
		for(int i=head2[u];i!=-1;i=edge[i].next){
			int v=edge[i].to;
			dist2[v]=max(dist2[v],dist2[u]);
			if(!vis2[v]){
				vis2[v]=1;
				q.push(v);
			}	
		}
	}
}
int main(){
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF){
		memset(head1,-1,sizeof(head1));
		memset(head2,-1,sizeof(head2));
		top=0;
		
		for(int i=1;i<=n;i++){
			scanf("%d",&dist1[i]);
			dist2[i]=dist1[i];
		}
		while(m--){
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			addedge(a,b);
		 	if(c==2)	addedge(b,a);		
		}
	
		SPFA1(1);
		SPFA2(n);
		int minn=0;
		for(int i=1;i<=n;i++){
			if(vis1[i]&&vis2[i])
			minn=max(minn,dist2[i]-dist1[i]);
		}
		printf("%d\n",minn);
	}
	return 0;
}