Fire Net HDU - 1045（二分匹配）

把每一列中相邻的 .  缩为一个点 作为二分图的左边

把每一行中相邻的  .  缩为一个点 作为二分图的右边

然后求最大匹配即可

这题用匈牙利足够了。。。然而。。我用了hk。。。有点大材小用的感觉///

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <algorithm>
#define mem(a, b) memset(a, b, sizeof(a))
using namespace std;
const int maxn = 10010, INF = 0x7fffffff;
int dx[maxn], dy[maxn], cx[maxn], cy[maxn], used[maxn];
int row[5][5], col[5][5];
int nx, ny, dis;
vector<int> G[40005];
char str[5][5];
int n;
int bfs()
{
    queue<int> Q;
    dis = INF;
    mem(dx, -1);
    mem(dy, -1);
    for(int i=1; i<=nx; i++)
    {
        if(cx[i] == -1)
        {
            Q.push(i);
            dx[i] = 0;
        }
    }
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        if(dx[u] > dis) break;
        for(int v=0; v<G[u].size(); v++)
        {
            int i = G[u][v];
            if(dy[i] == -1)
            {
                dy[i] = dx[u] + 1;
                if(cy[i] == -1) dis = dy[i];
                else
                {
                    dx[cy[i]] = dy[i] + 1;
                    Q.push(cy[i]);
                }
            }
        }
    }
    return dis != INF;
}

int dfs(int u)
{
    for(int v=0; v<G[u].size(); v++)
    {
        int i = G[u][v];
        if(!used[i] && dy[i] == dx[u] + 1)
        {
            used[i] = 1;
            if(cy[i] != -1 && dis == dy[i]) continue;
            if(cy[i] == -1 || dfs(cy[i]))
            {
                cy[i] = u;
                cx[u] = i;
                return 1;
            }
        }
    }
    return 0;
}

int hk()
{
    int res = 0;
    mem(cx, -1);
    mem(cy, -1);
    while(bfs())
    {
        mem(used, 0);
        for(int i=1; i<=nx; i++)
        {
            if(cx[i] == -1 && dfs(i)) res++;
        }
    }
    return res;
}

int main()
{
    while(cin>> n && n)
    {

        for(int i=0; i<100; i++) G[i].clear();
        mem(row, -1);
        mem(col, -1);
        nx = ny = 1;
        for(int i=0; i<n; i++)
        {
            cin>> str[i];
        }
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(str[i][j] == ‘.‘ && row[i][j] == -1)
                {
                    for(int k=j; str[i][k]==‘.‘ && k<n; k++)
                        row[i][k] = nx;
                    nx++;
                }
                if(str[j][i] == ‘.‘ && col[j][i] == -1)
                {
                    for(int k=j; str[k][i]==‘.‘ && k<n; k++)
                        col[k][i] = ny;
                    ny++;
                }
            }
        }
        nx -= 1, ny -= 1;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                if(str[i][j] == ‘.‘)
                    G[row[i][j]].push_back(nx + col[i][j]), G[nx + col[i][j]].push_back(row[i][j]);

        cout<< hk() <<endl;

    }

    return 0;
}

搜索写法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn = 6, INF = 0xfffffff;
typedef long long LL;
char str[maxn][maxn];
int vis[maxn][maxn];
int n, minn;
int check(int x,int y)
{
    for(int i=x-1; i>=0; --i)
    {
        if(vis[i][y])
            return 0;
        if(str[i][y] == ‘X‘)
            break;
    }
    for(int i=y-1; i>=0; --i)
    {
        if(vis[x][i])
            return 0;
        if(str[x][i] == ‘X‘)
            break;
    }
    return 1;
}

void dfs(int inx, int k)
{
    if(inx == n*n)
    {
        minn = max(k, minn);
        return;
    }

    int x = inx / n;
    int y = inx % n;
    if(str[x][y] == ‘.‘ && check(x,y))
    {
        vis[x][y] = 1;
        dfs(inx+1, k+1);
        vis[x][y] = 0;
    }
    dfs(inx+1, k);
}

int main()
{
    while(cin>>n && n)
    {
        minn = -INF;
        mem(vis,0);
        mem(str,0);
        for(int i=0;i<n;i++)
            cin>>str[i];
        dfs(0,0);
        cout<<minn<<endl;

    }

    return 0;
}

原文地址：https://www.cnblogs.com/WTSRUVF/p/9309488.html