二分图匹配入门题

板子（匈牙利算法，邻接矩阵）

const int MAXN=2e3+5;
int uN, vN;
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    for(int v=0; v<vN; v++)
        if(g[u][v] && !used[v])
        {
            used[v]=true;
            if(linker[v]==-1 || dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int res=0;
    memset(linker, -1, sizeof(linker));
    for(int u=0; u<uN; u++)
    {
        memset(used, 0, sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

HDU 1045

题意：给出一张图，给出空地‘.‘和隔板‘x’, 求放置最多满足条件的blockhouse，条件：垂直和水平方向上没有如果隔板隔开的话，只能放置一个house，有隔板话，隔板的之后（相对位置）的不用考虑。

题解：分别对每一行和每一列进行缩点（重新标号），两个相交的话就连边（其实就是把2个条件链接（行，列）到了一起）

#include <bits/stdc++.h>
using namespace std;
#define _for(i,a,b) for(int i=(a); i< (b); i++)
#define _rep(i,a,b) for(int i=(a); i<=(b); i++)

const int MAXN=2e2+5;
int uN, vN;
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    for(int v=0; v<vN; v++)
        if(g[u][v] && !used[v])
        {
            used[v]=true;
            if(linker[v]==-1 || dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int res=0;
    memset(linker, -1, sizeof(linker));
    for(int u=0; u<uN; u++)
    {
        memset(used, 0, sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

char Map[MAXN][MAXN];
int Mrow[MAXN][MAXN], Mcol[MAXN][MAXN];

int main()
{
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    //freopen("in.txt", "r", stdin);
    int n;
    while(cin>>n, n)
    {
        _for(i, 0, n) _for(j, 0, n) cin>>Map[i][j];
        uN=vN=0;
        int cu=0, cv=0;
        memset(Mrow, -1, sizeof(Mrow));
        memset(Mcol, -1, sizeof(Mcol));
        _for(i, 0, n) _for(j, 0, n)
        {
            if(Mrow[i][j]==-1 && Map[i][j]==‘.‘)
            {
                for(int k=j; k<n&&Map[i][k]==‘.‘; k++)
                    Mrow[i][k]=cu;
                uN=max(uN, ++cu);
            }
            if(Mcol[i][j]==-1 && Map[i][j]==‘.‘)
            {
                for(int k=i; k<n&&Map[k][j]==‘.‘; k++)
                    Mcol[k][j]=cv;
                vN=max(vN, ++cv);
            }
        }
        memset(g, 0, sizeof(g));
        _for(i, 0, n) _for(j, 0, n)
            if(Map[i][j]==‘.‘) g[Mrow[i][j]][Mcol[i][j]]=1;
        cout<<hungary()<<endl;
    }
    return 0;
}

HDU 2444

题意：给你一张图，你需要先判断这个图是不是二分图，然后在求其的最大匹配。

题解：用bfs对图进行染色，如果发现有相邻且同色的点那么就不是二分图。

#include <bits/stdc++.h>
using namespace std;

const int MAXN=200+5;
vector<int> G[MAXN];
int uN, linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    for(int i=0; i<G[u].size(); i++)
    {
        int v=G[u][i];
        if(!used[v])
        {
            used[v]=true;
            if(linker[v]==-1 || dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int res=0;
    memset(linker, -1, sizeof(linker));
    for(int u=1; u<=uN; u++)
    {
        memset(used, false, sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

int vis[MAXN];
bool check()
{
    queue<int> q;
    memset(vis, false, sizeof(vis));
    q.push(1); vis[1]=true;
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        for(int i=0; i<G[u].size(); i++)
        {
            int v=G[u][i];
            if(!vis[v])
            {
                if(vis[u]==1) vis[v]=2; else vis[v]=1;  //这里的染色别弄错了
                //if(vis[u]) vis[v]=2; else vis[v]=1;  WA
                q.push(v);
            }
            else if(vis[u]==vis[v])    return false;
        }
    }
    return true;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int n, m;
    while(cin>>n>>m)
    {
        uN=n;
        for(int i=0; i<=n; i++) G[i].clear();
        while(m--)
        {
            int u, v;
            cin>>u>>v;
            G[u].push_back(v);
            G[v].push_back(u);
        }
        if(check()) cout<<hungary()/2<<endl;
        else cout<<"No"<<endl;
    }
    return 0;
}

HDU 1281

题意：

　　小希和Gardon在玩一个游戏：对一个N*M的棋盘，在格子里放尽量多的一些国际象棋里面的“车”，并且使得他们不能互相攻击，这当然很简单，但是Gardon限制了只有某些格子才可以放，小希还是很轻松的解决了这个问题（见下图）注意不能放车的地方不影响车的攻击。 
　　所以现在Gardon想让小希来解决一个更难的问题，在保证尽量多的“车”的前提下，棋盘里有些格子是可以避开的，也就是说，不在这些格子上放车，也可以保证尽量多的“车”被放下。但是某些格子若不放子，就无法保证放尽量多的“车”，这样的格子被称做重要点。　Gardon想让小希算出有多少个这样的重要点，你能解决这个问题么。求多少个这样的点，和最大可以放下的车。

题解：把行看成x部，列看成y部，可以放车的点的坐标，即x，y可以构成一个匹配（和 HDU 1045一样的道理），每一行每一列只有一个车，相当于行和列在做匹配，在尝试接触g【x】【y】看求出的最大匹配是否发生变化。若发生变化，即是重要点。

#include <bits/stdc++.h>
using namespace std;

const int MAXN=100+5;
int uN, vN;
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    for(int v=0; v<vN; v++)
        if(g[u][v] && !used[v])
        {
            used[v]=true;
            if(linker[v]==-1 || dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int res=0;
    memset(linker, -1, sizeof(linker));
    for(int u=0; u<uN; u++)
    {
        memset(used, 0, sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

int main()
{
    int n, m, k, kase=0;
    while(cin>>n>>m>>k)
    {
        uN=n, vN=m;
        memset(g, 0, sizeof(g));
        while(k--)
        {
            int u, v;
            cin>>u>>v;
            g[u-1][v-1]=1;
        }
        int ans=hungary();
        int res=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            {
                if(g[i][j]) {
                    g[i][j]=0;
                    if(ans>hungary()) res++;
                    g[i][j]=1;
                }
            }
         printf("Board %d have %d important blanks for %d chessmen.\n",++kase, res, ans);
    }
    return 0;
}

HDU 2819

题意：可交换任意两行或任意两列，最终是主对角线上全为1，输出交换过程（如果可以的话），否则-1

题解：如果可行的话，一定可以只交换列或只交换行得到，因为不管怎么交换，原先在同一行的始终在同一行，原先在同一列的始终在同一列。用行列构图（和上面的一样），如果a[i][j]=1，则加边，求最大匹配就行。

　　　输出交换的路径的时候，注意这个交换中的行列是指在原图的中的位置。

　　　注意：解除流绑定后，就要不要用printf了，用printf就不要接触流绑定，会WA

#include <bits/stdc++.h>
using namespace std;

const int MAXN=200+5;
int uN, vN;
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    for(int v=0; v<vN; v++)
        if(g[u][v] && !used[v])
        {
            used[v]=true;
            if(linker[v]==-1 || dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary()
{
    int res=0;
    memset(linker, -1, sizeof(linker));
    for(int u=0; u<uN; u++)
    {
        memset(used, 0, sizeof(used));
        if(dfs(u)) res++;
    }
    return res;
}

int A[MAXN*MAXN], B[MAXN*MAXN];

int main()
{
    //freopen("in.txt", "r", stdin);
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n;
    while(cin>>n)
    {
        uN=vN=n;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                cin>>g[i][j];
        if(hungary()<n){
            cout<<"-1"<<endl;
            continue;
        }
        int cnt=0;
        for(int j, i=0; i<uN; i++)  //记录匹配过程中的匹配路径
        {
            for(j=0; j<vN&&linker[j]!=i; j++) ;
            if(i!=j){               //交换i，j。这个交换中的行列是指在原图的中的位置
                A[cnt]=i, B[cnt++]=j;
                swap(linker[i], linker[j]);
            }
        }
        cout<<cnt<<endl;
        for(int i=0; i<cnt; i++)
            printf("C %d %d\n", A[i]+1, B[i]+1);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Yokel062/p/11330059.html

时间： 2024-08-07 10:43:05

二分图匹配入门题的相关文章

【二分图匹配入门专题1】D - Matrix hdu2119【最小顶点覆盖】

Give you a matrix(only contains 0 or 1),every time you can select a row or a column and delete all the '1' in this row or this column . Your task is to give out the minimum times of deleting all the '1' in the matrix. InputThere are several test case

【二分图匹配入门专题1】H - Marriage Media light oj 1184【二分图最大匹配】

You run a marriage media. You take some profiles for men and women, and your task is to arrange as much marriages as you can. But after reading their bio-data you have found the following criteria. No man will marry a woman if their height gap is gre

【二分图匹配入门专题1】F - COURSES poj1469【最大匹配--匈牙利算法模板题】

Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions: every stude

【二分图匹配入门专题1】I - Hiding Gold light oj 1152【二分图匹配】-------------------我是终于不那么水的水题分割线------------------------

You are given a 2D board where in some cells there are gold. You want to fill the board with 2 x 1 dominoes such that all gold are covered. You may use the dominoes vertically or horizontally and the dominoes may overlap. All you have to do is to cov

【二分图匹配入门专题1】E - Air Raid hdu1151【最小路径覆盖】

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town's streets you can never reach the same intersection i.e. the town's

【二分图匹配入门专题1】K - Going Home hdu1533【km匹配】

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, unt

【二分图匹配入门专题1】M - Cyclic Tour hdu1853【km算法--判断自重边】

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total len

【二分图匹配入门专题1】 N - Special Fish hdu3395 【km算法】【春风十里，都不如tle~~~】

There is a kind of special fish in the East Lake where is closed to campus of Wuhan University. It's hard to say which gender of those fish are, because every fish believes itself as a male, and it may attack one of some other fish who is believed to

【二分图匹配入门专题1】B - Girls and Boys hdu1068【最大独立集】

the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary to find out the maximu