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=1;i<=n;i++)
        for(int j=0;j<=k+2;j++)
            dist[i][j]=inf;
    queue<Pair>Q;
    Q.push(make_pair(1,0));
    dist[1][0]=0;
    mark[1][0]=true;
    while(!Q.empty()){
        Pair pp=Q.front();
        Q.pop();
        int u=pp.first,c=pp.second;
        mark[u][c]=false;
        for(int i=0;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=0;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+2);
        while(m--){
            scanf("%d%d%d%d",&u,&v,&w,&c);
            map[u].push_back(Edge(v,w,c));
        }
        if(SPFA()){
            printf("%d
",MIN);
        }else
            puts("-1");
    }
    return 0;
}

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=1,p.dd=0,p.cost=0,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=0;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+2);
        while(m--){
            scanf("%d%d%d%d",&u,&v,&w,&c);
            map[u].push_back(Edge(v,w,c));
        }
        if(Dijkstra()){
            printf("%d
",MIN);
        }else
            puts("-1");
    }
    return 0;
}