HDU-4185-Oil Skimming [二分匹配]

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題目傳送門

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題意:一個地圖由’.’和’#’組成，問有多少個不重疊的’##’。

思路:先將所有的’#’hash，然後建二分圖，用匈牙利算法算最大匹配日數，因爲邊建了兩次，所以結果除以二。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

struct node{
    int to, next;
}edge[720100];
char mp[610][610];
int a[610][610];
int n, cnt, head[360100];
int num;
int match[360100];
bool used[360100];
void add(int u, int v)
{
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
bool Dfs(int x)
{
    for (int i = head[x]; i; i = edge[i].next)
    {
        int v = edge[i].to;
        if (!used[v])
        {
            used[v] = true;
            if (!match[v] || Dfs(match[v]))
            {
                match[v] = x;
                return true;
            }
        }
    }
    return false;
}
int Matching()
{
    int ans = 0;
    memset(match,0,sizeof(match));
    for (int i = 1; i <= num; i++)
    {
        memset(used, false, sizeof(used));
        if (Dfs(i))
        {
            ans++;
        }
    }
    return ans;
}
int main(void)
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);

    int T, cas = 1;
    scanf("%d", &T);
    while (T--)
    {
        memset(head, 0, sizeof(head));
        memset(a,0,sizeof(a));
        memset(mp, 0, sizeof(mp));
        cnt = 1;
        num = 1;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf(" %s", mp[i]);
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                if (mp[i][j]=='#')
                    a[i][j] = num++;
        num--;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (a[i][j])
                {
                    if (i-1>=0 && a[i-1][j])
                    {
                        add(a[i][j], a[i-1][j]);
                    }
                    if (i+1<n && a[i+1][j])
                    {
                        add(a[i][j], a[i+1][j]);
                    }
                    if (j+1<n && a[i][j+1])
                    {
                        add(a[i][j], a[i][j+1]);
                    }
                    if (j-1>=0 && a[i][j-1])
                    {
                        add(a[i][j], a[i][j-1]);
                    }
                }
            }
        }
        int ans = Matching();
        printf("Case %d: %d\n", cas++, ans/2);
    }
    return 0;
}