poj 1724(有限制的最短路)

题目链接：http://poj.org/problem?id=1724

思路：

有限制的最短路，或者说是二维状态吧，在求最短路的时候记录一下花费即可。一开始用SPFA写的，900MS险过啊，然后改成Dijkstra+priority_queue,60MS,orz.

SPFA:

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<vector>
 using namespace std;
 #define MAXN 104
 #define inf 1<<30
 typedef pair<int,int>Pair;

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

 vector<vector<Edge> >map;
 int dist[MAXN][MAXN*MAXN];//在点i花费j的最短路径
 bool mark[MAXN][MAXN*MAXN];
 int k,n,m,MIN;

 bool SPFA()
 {
     memset(mark,false,sizeof(mark));
     for(int i=;i<=n;i++)
         for(int j=;j<=k+;j++)
             dist[i][j]=inf;
     queue<Pair>Q;
     Q.push(make_pair(,));
     dist[][]=;
     mark[][]=true;
     while(!Q.empty()){
         Pair pp=Q.front();
         Q.pop();
         int u=pp.first,c=pp.second;
         mark[u][c]=false;
         for(int i=;i<map[u].size();i++){
             int v=map[u][i].v;
             int w=map[u][i].w;
             int cc=map[u][i].c+c;
             if(cc>k)continue;
             if(dist[u][c]+w<dist[v][cc]){
                 dist[v][cc]=dist[u][c]+w;
                 if(!mark[v][cc]){
                     mark[v][cc]=true;
                     Q.push(make_pair(v,cc));
                 }
             }
         }
     }
     MIN=inf;
     for(int i=;i<=k;i++)MIN=min(MIN,dist[n][i]);
     return MIN<inf;
 }        

 int main()
 {
     int u,v,w,c;
     while(~scanf("%d%d%d",&k,&n,&m)){
         map.clear();map.resize(n+);
         while(m--){
             scanf("%d%d%d%d",&u,&v,&w,&c);
             map[u].push_back(Edge(v,w,c));
         }
         if(SPFA()){
             printf("%d\n",MIN);
         }else
             puts("-1");
     }
     return ;
 }

Dijkstra+priority_queue:

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<vector>
 using namespace std;
 #define MAXN 104
 #define inf 1<<30
 typedef pair<int,int>Pair;

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

 struct Node{
     int u,dd,cost;
     bool operator < (const Node &p) const {
         if(p.dd==dd)return p.cost<cost;
         return p.dd<dd;
     }
 };

 vector<vector<Edge> >map;
 int k,n,m,MIN;

 bool Dijkstra()
 {
     priority_queue<Node>Q;
     Node p,q;
     p.u=,p.dd=,p.cost=,MIN=inf;
     Q.push(p);
     while(!Q.empty()){
         p=Q.top();
         Q.pop();
         int u=p.u,dd=p.dd,cost=p.cost;
         if(u==n){ MIN=dd;break; }
         for(int i=;i<map[u].size();i++){
             int v=map[u][i].v,w=map[u][i].w,c=map[u][i].c;
             if(cost+c<=k){
                 q.u=v;q.dd=dd+w;q.cost=cost+c;
                 Q.push(q);
             }
         }
     }
     return MIN<inf;
 }

 int main()
 {
     int u,v,w,c;
     while(~scanf("%d%d%d",&k,&n,&m)){
         map.clear();map.resize(n+);
         while(m--){
             scanf("%d%d%d%d",&u,&v,&w,&c);
             map[u].push_back(Edge(v,w,c));
         }
         if(Dijkstra()){
             printf("%d\n",MIN);
         }else
             puts("-1");
     }
     return ;
 }