free

问题 J: free

时间限制: 2 Sec  内存限制: 128 MB
提交: 4  解决: 3
[提交] [状态] [命题人:admin]

题目描述

Your are given an undirect connected graph.Every edge has a cost to pass.You should choose a path from S to T and you need to pay for all the edges in your path. However, you can choose at most k edges in the graph and change their costs to zero in the beginning. Please answer the minimal total cost you need to pay.

 

输入

The first line contains five integers n,m,S,T,K.
For each of the following m lines, there are three integers a,b,l, meaning there is an edge that costs l between a and b.
n is the number of nodes and m is the number of edges.

 

输出

An integer meaning the minimal total cost.

 

样例输入

复制样例数据

3 2 1 3 1
1 2 1
2 3 2

样例输出

1

 

提示

1≤n,m≤103,1≤S,T,a,b≤n,0≤k≤m,1≤l≤106.
Multiple edges and self loops are allowed.

#include<bits/stdc++.h>
#include<iostream>

#define ll long long
using namespace std;
const int p = 1e9 + 7;
const int maxn = 2e3 + 10;

struct path{
    int to,nx,w;
}o[maxn];

struct atom{
    int ver,val,cost;
    atom(int _ver,int _val,int _cost):ver(_ver),val(_val),cost(_cost){}
    bool operator<(const atom& cur)const {
        return val>cur.val;
    }
};

int n,m,S,T,k,tot,head[maxn],vis[maxn][maxn];
ll dp[maxn][maxn];

inline void add_edge(int u,int v,int w){
    o[++tot].to=v;
    o[tot].w=w;
    o[tot].nx=head[u];
    head[u]=tot;
}

inline void dij(){
    for(register int i=1;i<=n;++i){
        for(register int j=0;j<=k;++j){
            dp[i][j]=-1;
        }
    }
    priority_queue<atom>q;
    q.push(atom(S,0,k));
    dp[S][k]=0;
    while(!q.empty()){
        atom cur=q.top();
        q.pop();
        int point=cur.ver,cnt=cur.cost;
        if(vis[point][cnt])continue;
        vis[point][cnt]=1;
        for(register int i=head[point];i;i=o[i].nx){
            int to=o[i].to;
            if(cnt>0&&(dp[to][cnt-1]==-1||dp[point][cnt]<dp[to][cnt-1])){
                dp[to][cnt-1]=dp[point][cnt];
                q.push(atom(to,dp[to][cnt-1],cnt-1));
            }
            if(dp[to][cnt]==-1||dp[to][cnt]>dp[point][cnt]+o[i].w){
                dp[to][cnt]=dp[point][cnt]+o[i].w;
                q.push(atom(to,dp[to][cnt],cnt));
            }
        }
    }
}


int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt","r",stdin);
#endif
    scanf("%d%d%d%d%d",&n,&m,&S,&T,&k);
    for(register int i=1;i<=m;++i){
        int a,b,l;
        scanf("%d%d%d",&a,&b,&l);
        add_edge(a,b,l);
        add_edge(b,a,l);
    }
    dij();
    ll res=INT_MAX/2;
    for(register int i=0;i<=k;++i){
        if(dp[T][i]!=-1)res=min(res,dp[T][i]);
    }
    printf("%lld\n",res);
    return 0;
}