2019牛客暑期多校訓練營（第四場）- J - free

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.

樣例演示：

輸入：

3 2 1 3 1
1 2 1
2 3 2

輸出：

1

數據範圍：

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

題目大意：給定一張無向圖，起點爲S，終點爲T，求S-T的最短路，其中可以有k次機會把一條路徑的權值變爲0。

解題思路：分層最短路

AC代碼：

#pragma GCC optimize(3)
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<array>
#include<string>
#include<cstring>
#include<cmath>
#include<cassert>
#include<cstdlib>
#include<utility>
#include<iterator>
using namespace std;
typedef long long ll;
const int maxn  =  1005;
int n,m,s,t,k;
struct  edge
{
    int to,cost;
    edge(){}
    edge(int to,int cost):to(to),cost(cost){}
};
struct  node
{
    int v,w,cnt;
    node(){}
    node(int v,int w,int cnt):v(v),w(w),cnt(cnt){}
    bool operator<(const node&ano)const{return w>ano.w;}
};
int dis[maxn][maxn];
bool vis[maxn][maxn];
vector<edge> G[maxn];
priority_queue<node> pq;
void addG(int u,int v,int w)
{
    G[u].push_back(edge(v,w));
    G[v].push_back(edge(u,w));
}
void dijkstra()
{
    memset(dis,0x3f,sizeof(dis));
    memset(vis,false,sizeof(vis));
    dis[s][0] = 0;
    pq.push(node(s,0,0));
    while(!pq.empty())
    {
        const node top = pq.top();
        pq.pop();
        int u = top.v;
        int nowcnt = top.cnt;
        if(vis[u][nowcnt]==true)continue;
        vis[u][nowcnt] = true;
        for(int i = 0;i<(int)(G[u].size());i++)
        {
            int v = G[u][i].to;
            int w = G[u][i].cost;
            //可以用到免費機會
            if(nowcnt<k&&dis[v][nowcnt+1]>dis[u][nowcnt]&&!vis[v][nowcnt+1])
            {    
                  dis[v][nowcnt+1]=dis[u][nowcnt];
                  pq.push(node(v,dis[v][nowcnt+1],nowcnt+1));
            }
            //不可以用免費機會
            if(dis[v][nowcnt]>dis[u][nowcnt]+w&&!vis[v][nowcnt])
            {
                dis[v][nowcnt]=dis[u][nowcnt]+w;
                pq.push(node(v,dis[v][nowcnt],nowcnt));
            }
        }

    }
}
int main()
{
    ios::sync_with_stdio(false);cin.tie(0);
    cin>>n>>m>>s>>t>>k;
    while(m--)
    {
        int u,v,w;
        cin>>u>>v>>w;
        addG(u,v,w);
    }
    int ans = 0x3f3f3f3f;
    dijkstra();
    for(int i = 0;i<=k;i++)
    {
        ans = min(ans,dis[t][i]);
    }
    cout<<ans<<endl;
    return 0;
}