牛客多校__free

鏈接：https://ac.nowcoder.com/acm/contest/884/J
來源：牛客網
 

時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 524288K，其他語言1048576K
64bit IO Format: %lld

題目描述

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.

輸入描述:

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.

輸出描述:

An integer meaning the minimal total cost.

示例1

輸入

複製

3 2 1 3 1
1 2 1
2 3 2

輸出

複製

1

題意：在一個圖中讓k條邊的權值變爲0，然後輸出兩點之間最短距離

題解：在跑dijk最短路的同時，進行dp，dp[i][j]表示在第i個點時，讓j條邊變爲0的最短距離，這樣就能寫出遞推公式

dp[v][j]=min(dp[v][j],dp[u][j-1])

dp[v][j]=min(dp[v][j],dp[u][j]+edge[i].c)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int,int>P;

const int maxn =1e3+9;

ll dp[maxn][maxn];

int n,m,s,t,k;
int tot;
int head[maxn];

struct rt{
    int v,next;
    ll c;
}edge[maxn*100];

void add_edge(int u,int v,ll c){
    edge[tot].v=v;
    edge[tot].c=c;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void dijk(int s){
    priority_queue<P,vector<P>,greater<P> >que;
    memset(dp,0x3f,sizeof(dp));
    for(int i=0;i<=k;i++)dp[s][i]=0;
    que.push(P(0,s));
    while(!que.empty()){
        P p=que.top(); que.pop();
        int u=p.second;
        if(dp[u][0]<p.first)continue;
//        cout<<u<<endl;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            int flag=0;
            for(int j=1;j<=k;j++){
                if(dp[v][j]>dp[u][j-1])
                {
                    dp[v][j]=dp[u][j-1];
                    flag=1;
                }
            }
            for(int j=0;j<=k;j++){
                if(dp[v][j]>dp[u][j]+edge[i].c){
                    dp[v][j]=dp[u][j]+edge[i].c;
                    flag=1;
                }
            }
            if(flag){
                que.push(P(dp[v][0],v));
            }
        }
    }
}


int main(){
    memset(head,-1,sizeof(head));
    scanf("%d%d%d%d%d",&n,&m,&s,&t,&k);
    int u,v;
    ll c;
    for(int i=0;i<m;i++){
        scanf("%d%d%lld",&u,&v,&c);
        add_edge(u,v,c);
        add_edge(v,u,c);
    }
    dijk(s);
    printf("%lld\n",dp[t][k]);
    return 0;
}