hdu 5352 MZL's City 【二分图】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5352

题意：
给你n,m,k
表示n个建筑 m次操作，修复操作每次最多修复k个建筑。
有三种操作
1.修复x点周围建筑（<=k）
2.x,y建筑间建边
3.x,y间删边
修复建筑时候拆点建图。反着求最大匹配，保证字典序最小。

代码：

#include <stdio.h>  
#include <ctime>  
#include <math.h>  
#include <limits.h>  
#include <complex>  
#include <string>  
#include <functional>  
#include <iterator>  
#include <algorithm>  
#include <vector>  
#include <stack>  
#include <queue>  
#include <set>  
#include <map>  
#include <list>  
#include <bitset>  
#include <sstream>  
#include <iomanip>  
#include <fstream>  
#include <iostream>  
#include <ctime>  
#include <cmath>  
#include <cstring>  
#include <cstdio>  
#include <time.h>  
#include <ctype.h>  
#include <string.h>  
#include <assert.h>  
#pragma comment (linker, "/STACK:102400000,102400000")

using namespace std;

template <class T>
inline bool scan_d(T &ret)
{
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0; //EOF  
    while (c != ‘-‘ && (c<‘0‘ || c>‘9‘)) c = getchar();
    sgn = (c == ‘-‘) ? -1 : 1;
    ret = (c == ‘-‘) ? 0 : (c - ‘0‘);
    while (c = getchar(), c >= ‘0‘&&c <= ‘9‘) ret = ret * 10 + (c - ‘0‘);
    ret *= sgn;
    return 1;
}

const int MAXN = 100100;
const int N = 210;
// 200 500 200
int n, m, k;

int mp[N][N];
int book[N];
int match[N];
int path[N];
int len;
int vis[N];
vector<int> g[MAXN];
int ans[N*3], res;
int cc;

void get_path(int u)
{
    vis[u] = 1;
    path[len++] = u;
    for (int i = 1; i <= n; i++)
    {
        if (!vis[i] && mp[u][i] == 1)
        {
            get_path(i);
        }
    }
}

bool dfs(int u)
{
    int i;
    for (i = 0; i < g[u].size(); i++)
    {
        int v = g[u][i];
        if (book[v] == 0)
        {
            book[v] = 1;
            if (match[v] == 0 || dfs(match[v]))
            {
                match[v] = u;
                return true;
            }
        }
    }
    return false;
}

void hungry()
{
    for (int i = cc-1; i >= 0; i--)//保证字典序小 优先匹配后面的
    {
        for (int j = i*k; j < i*k + k; j++)
        {
            memset(book, 0, sizeof(book));
            if (dfs(j))
                ans[i]++;
        }
        res += ans[i];
    }
}

void init()
{
    cc = 0;
    res = 0;
    for (int i = 0; i < 100010; i++) g[i].clear();
    memset(mp, 0, sizeof(mp));
    memset(match, 0, sizeof(match));
    memset(ans, 0, sizeof(ans));
}

int main()
{
    int t;
    scan_d(t);
    while (t--)
    {
        scan_d(n);
        scan_d(m);
        scan_d(k);
        init();
        while(m--)
        {
            int p, q;
            int x, y;
            scan_d(p);
            if (p == 1)
            {
                scan_d(x);
                memset(vis, 0, sizeof(vis));
                len = 0;
                get_path(x);
                for (int i = 0; i < len; i++)
                {
                    for (int j = cc*k; j < (cc + 1)*k; j++)
                        g[j].push_back(path[i]);
                }
                cc++;
            }
            else if (p == 2)
            {
                scan_d(x);
                scan_d(y);
                mp[x][y] = mp[y][x] = 1;
            }
            else if (p == 3)
            {
                scan_d(q);
                while (q--)
                {
                    scan_d(x);
                    scan_d(y);
                    mp[x][y] = mp[y][x] = 0;
                }
            }
        }

        hungry();
        printf("%d\n", res);
        for (int i = 0; i < cc - 1; i++)
            printf("%d ", ans[i]); printf("%d\n", ans[cc - 1]);

    }
    return 0;
}