poj 3159(差分约束经典题)

题目链接：http://poj.org/problem?id=3159思路：题目意思很简单，都与给定的条件dist[b]-dist[a]<=c,求dist[n]-dist[1]的最大值，显然这是差分约束的经典题，条件可以转化为dist[b]<=dist[a]+c,于是a->b直接连边，边权值为c,从而题目转化为图上求1->n的最短路，看了一下数据，30000个点，150000条边，果断用Dijkstra+priority_queue,1300MS+险过，orz.

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<vector>
 using namespace std;
 #define MAXN 30030
 #define inf 1<<30
 typedef pair<int,int>Pair;

 struct Edge{
     int v,w;
     Edge(int vv,int ww):v(vv),w(ww){}
 };

 int dist[MAXN];
 bool mark[MAXN];
 int n,m;

 vector<vector<Edge> >map;

 int Dijkstra(int u)
 {
     memset(mark,false,(n+)*sizeof(bool));
     for(int i=;i<=n;i++)dist[i]=inf;
     dist[u]=;
     priority_queue<Pair,vector<Pair>,greater<Pair> >Q;
     Q.push(make_pair(,u));
     while(!Q.empty()){
         Pair pp=Q.top();
         Q.pop();
         int dd=pp.first,u=pp.second;
         if(mark[u])continue;
         mark[u]=true;
         for(int i=;i<map[u].size();i++){
             int v=map[u][i].v,w=map[u][i].w;
             if(mark[v])continue;
             if(dd+w<dist[v]){
                 dist[v]=dd+w;
                 Q.push(make_pair(dist[v],v));
             }
         }
     }
     return dist[n];
 }

 int main()
 {
     int u,v,w;
     scanf("%d%d",&n,&m);
     map.clear();map.resize(n+);
     while(m--){
         scanf("%d%d%d",&u,&v,&w);
         map[u].push_back(Edge(v,w));
     }
     printf("%d\n",Dijkstra());
     return ;
 }