[CF852D] Exploration plan

Posted 2020-11-16 lslzf

问题描述

The competitors of Bubble Cup X gathered after the competition and discussed what is the best way to get to know the host country and its cities.

After exploring the map of Serbia for a while, the competitors came up with the following facts: the country has V cities which are indexed with numbers from 1 to V, and there are E bi-directional roads that connect the cites. Each road has a weight (the time needed to cross that road). There are N teams at the Bubble Cup and the competitors came up with the following plan: each of the N teams will start their journey in one of the V cities, and some of the teams share the starting position.

They want to find the shortest time T, such that every team can move in these T minutes, and the number of different cities they end up in is at least K (because they will only get to know the cities they end up in). A team doesn‘t have to be on the move all the time, if they like it in a particular city, they can stay there and wait for the time to pass.

Please help the competitors to determine the shortest time T so it‘s possible for them to end up in at least K different cities or print -1 if that is impossible no matter how they move.

Note that there can exist multiple roads between some cities.

输入格式

The first line contains four integers: V, E, N and K (1?≤??V??≤??600,? 1??≤??E??≤??20000,? 1??≤??N??≤??min(V,?200),? 1??≤??K??≤??N), number of cities, number of roads, number of teams and the smallest number of different cities they need to end up in, respectively.

The second line contains N integers, the cities where the teams start their journey.

Next E lines contain information about the roads in following format: Ai Bi Ti (1?≤?Ai,?Bi?≤?V,? 1?≤?Ti?≤?10000), which means that there is a road connecting cities Ai and Bi, and you need Ti minutes to cross that road.

输出格式

Output a single integer that represents the minimal time the teams can move for, such that they end up in at least K different cities or output -1 if there is no solution.

If the solution exists, result will be no greater than 1731311.

样例输入

6 7 5 4
5 5 2 2 5
1 3 3
1 5 2
1 6 5
2 5 4
2 6 7
3 4 11
3 5 3

样例输出

3

样例解释

Three teams start from city 5, and two teams start from city 2. If they agree to move for 3 minutes, one possible situation would be the following: Two teams in city 2, one team in city 5, one team in city 3 , and one team in city 1. And we see that there are four different cities the teams end their journey at.

题目大意

给定一个 v个点 e条边的带权无向图，在图上有 n个人，第 i个人位于点 xi，一个人通过一条边需要花费这条边的边权的时间。

现在每个人可以自由地走。求最短多少时间后满足结束后有人的节点数 ≥ m

解析

观察到最后的答案就是走过的最长时间。那么，这就变成了一个最大值最小的问题，可以用二分答案解决。

二分需要的时间mid。因为最后是至少m做城市有人，所以不妨当做是用m个人去匹配m座城市，那么就变成了一个二分图匹配问题。对于每个人，向他所在的城市在mid时间内可以到达的城市连边，这可以用Floyd求出两两最短路得到。然后二分图匹配，如果匹配数大于等于m说明可行。

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 1202
using namespace std;
int v,e,n,m,i,j,k,dis[N][N],g[N][N],pos[N],match[N];
bool vis[N];
int read()
{
    char c=getchar();
    int w=0;
    while(c<'0'||c>'9') c=getchar();
    while(c<='9'&&c>='0'){
        w=w*10+c-'0';
        c=getchar();
    }
    return w;
}
bool dfs(int x)
{
    for(int y=1;y<=v;y++){
        if(g[x][y]&&!vis[y]){
            vis[y]=1;
            if(!match[y]||dfs(match[y])){
                match[y]=x;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    memset(match,0,sizeof(match));
    int ans=0;
    for(int i=1;i<=n;i++){
        memset(vis,0,sizeof(vis));
        if(dfs(i)) ans++;
    }
    return ans;
}
bool check(int x)
{
    memset(g,0,sizeof(g));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=v;j++){
            if(dis[pos[i]][j]<=x) g[i][j]=1;
        }
    }
    int ans=hungary();
    return (ans>=m);
}
int main()
{
    v=read();e=read();n=read();m=read();
    for(i=1;i<=n;i++) pos[i]=read();
    memset(dis,0x3f,sizeof(dis));
    for(i=1;i<=v;i++) dis[i][i]=0;
    for(i=1;i<=e;i++){
        int u=read(),v=read(),w=read();
        dis[u][v]=dis[v][u]=min(dis[u][v],w);
    }
    for(k=1;k<=v;k++){
        for(i=1;i<=v;i++){
            for(j=1;j<=v;j++) dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
        }
    }
    int l=0,r=1731311,mid,ans=-1;
    while(l<=r){
        mid=(l+r)/2;
        if(check(mid)){
            ans=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    printf("%d
",ans);
    return 0;
}