POJ 1724 最短路费用限制

迪杰斯塔拉裸题

最大花费

n个点

m条有向边

起点终点 路径长度 路径花费

问：在花费限制下，最短路径的长度

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cctype>
#include <queue>
#include <stdlib.h>
#include <cstdlib>
#include <math.h>
#include <set>
#include <vector>
#define inf 107374182
#define N 101
#define M 10001
#define ll int
using namespace std;
inline ll Max(ll a,ll b){return a>b?a:b;}
inline ll Min(ll a,ll b){return a<b?a:b;}
struct Edge{
	int f,t,d,w;
	int nex;
}edge[M];
int head[N],edgenum;
void addedge(int u,int v,int d,int w){
	Edge E={u,v,d,w,head[u]};
	edge[edgenum]=E;
	head[u]=edgenum++;
}

struct node{
	int to,dd,use;
	node(int a=0,int c=0,int b=0):to(a),dd(c),use(b){}
	bool operator<(const node&a)const{
		if(a.dd==dd)return a.use<use;
		return a.dd<dd;
	}
};
int n,maxcost,dis[N];
void spfa(int s,int e){
	int i;
	for(i=1;i<=n;i++)dis[i]=inf;
	dis[s]=0;

	priority_queue<node>q; while(!q.empty())q.pop();
	q.push(node(s,0,0));
	while(!q.empty())
	{
		node temp=q.top(); q.pop();
		int u=temp.to,nowcost=temp.use,d=temp.dd;
		if(u==e)return;
		for(i=head[u];i!=-1;i=edge[i].nex)
		{
			int v=edge[i].t;
			if(nowcost+edge[i].w<=maxcost)
			{
				if(dis[v]>d+edge[i].d)
				dis[v]=d+edge[i].d;
				q.push(node(v,d+edge[i].d,nowcost+edge[i].w));
			}
		}
	}
}
int main()
{
	int i,m,u,v,d,w;
	while(~scanf("%d",&maxcost)){
		scanf("%d%d",&n,&m);
		memset(head,-1,sizeof(head));	edgenum=0;

		while(m--){
			scanf("%d %d %d %d",&u,&v,&d,&w);
			addedge(u,v,d,w);
		}

		spfa(1,n);
		if(dis[n]==inf)dis[n]=-1;
		printf("%d\n",dis[n]);
	}
	return 0;
}