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 ;

 }