Gym - 101670J Punching Power（最大獨立集）

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題目：

The park management finally decided to install some popular boxing machines at various strategic places in the park. In fact, to compensate for the previous lack of machines, they decided to install as many machines as possible. Surprisingly enough, the park is not going to be choked with new machines because there are some quite serious legal limitations regarding the locations of the machines. The management has marked all possible boxing machine locations and their respective coordinates on the park plan. Additionally, they also have to respect manufacturer security rule: The distance between any two boxing machines has to be at least 1.3 meters.

Help the management to establish the maximum possible number of boxing machines which can be installed in the park.

Input Specification：

There are several test cases. Each case starts with a line containing one integer N which specifies the number of possible boxing machine locations in the park (1 ≤ N ≤ 2000). Next, there are N lines representing the location coordinates, each line describes one location by a pair of integer coordinates in meters. All locations in one test case are unique. Each coordinate is non-negative and less than or equal to 109 .

You are guaranteed that all locations form a single connected group, that is, it is possible to start in any location and reach any other location by a sequence of steps, each of which changes exactly one coordinate by 1, without leaving the area suitable for placing boxing machines.

Output Specification:

For each test case, print a single line with one integer representing the maximum number of boxing machines which can be installed in the park.

思路：

先吐槽一下：

The distance between any two boxing machines has to be at least 1.3 meters.這句話難道沒有用？？？？

建圖的規則是根據each of which changes exactly one coordinate by 1這句話來的,,,,,,,,,,直接自閉。

首先建好圖，然後求最大獨立集就ok了，最大獨立集 = 點的個數 - 最大匹配數。

總結這道題被卡的原因：

對匈牙利演算法的理解太淺顯！

讀題不精！

代碼：

 

#include <bits/stdc++.h>#define inf 0x3f3f3f3fusing namespace std;typedef long long ll;const int maxn = 3000;struct Node{    int x,y;} node[maxn];vector<int>v[maxn];int n,vis[maxn],linker[maxn];bool dfs(int u){    for(int i = 0; i<v[u].size(); i++)//每個與u相連的點    {        int _v = v[u][i];//放進交替路中        if(!vis[_v])        {            vis[_v] = 1;            if(linker[_v]==0 || dfs(linker[_v]))//是未匹配點，說明該交替路是增廣路徑，交換路徑            {                linker[_v] = u;                linker[u] = _v;                return true;            }        }    }    return false;}int match(){    int res = 0;    memset(linker,0,sizeof(linker));    for(int i = 0; i<n; i++)    {        memset(vis,0,sizeof(vis));        if(linker[i]==0 && dfs(i))            res++;    }    return res;}int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i = 1; i<=n; i++)        {            scanf("%d%d",&node[i].x,&node[i].y);        }        for(int i = 1; i<=n; i++)        {            v[i].clear();            for(int j = 1; j<=n; j++)            {                if(abs(node[i].x-node[j].x)+abs(node[i].y-node[j].y)==1)                {                    v[i].push_back(j);                }            }        }        int ans = match();        //cout<<"ans: "<<ans<<endl;        printf("%d\n",n-ans);    }    return 0;}

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Gym - 101670J Punching Power（最大獨立集）