ZOJ1516_Uncle Tom's Inherited Land_二分图（匈牙利算法）

smmrSangria

于 2017-09-03 09:42:04 发布

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分类专栏：二分图文章标签： zoj Uncle Toms Inherited 二分图匈牙利算法

本文链接：https://blog.youkuaiyun.com/Anna__1997/article/details/77815941

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本文介绍了一种通过构建二分图并使用匈牙利算法来解决01矩阵中寻找最多可分割成1*2矩阵数量的问题。核心思路在于将相邻的0视为节点，并在它们之间建立边，最终通过最大匹配数确定分割的数量。

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题目大意
给一个01矩阵，求最对能分割出几块由0构成的1 * 2的矩阵
思路
抽象出二分图，相邻的0之间建边，匈牙利算法求最大匹配数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#define INF 0x3f3f3f3f
#define rep0(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)
#define rep_1(i, n) for (int i = n; i > 0; i--)
#define MAX(x, y) (((x) > (y)) ? (x) : (y))
#define MIN(x, y) (((x) < (y)) ? (x) : (y))
#define mem(x, y) memset(x, y, sizeof(x))
#define MAXN 110

using namespace std;
struct Edge
{
    int from, to;
    Edge(int f, int t) : from(f), to(t) {}
};
vector<Edge> edges;
vector<int> g[MAXN * MAXN];
int land[MAXN][MAXN], matching[MAXN * MAXN], check[MAXN * MAXN];
void addEdge(int from, int to)
{
    edges.push_back(Edge(from, to));
    g[from].push_back(edges.size() - 1);
    edges.push_back(Edge(to, from));
    g[to].push_back(edges.size() - 1);
}
int n, m, k;
bool dfs(int u)
{
    for (int i = 0; i < g[u].size(); i++)
    {
        int v = edges[g[u][i]].to;
        if (check[v])
            continue;
        check[v] = 1;
        if (matching[v] == -1 || dfs(matching[v]))
        {
            matching[v] = u;
            matching[u] = v;

            return true;
        }
    }
    return false;
}
int hungarian()
{
    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = (i - 1) % 2 + 1; j <= m; j += 2)
        {
            if (land[i][j] == 1)
                continue;
            mem(check, 0);

            if (dfs((i - 1) * m + j))
                ans++;
        }
    }
    return ans;
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE

    while (~scanf("%d %d %d", &n, &m, &k) && n)
    {
        mem(land, 0);
        mem(matching, -1);
        int a, b;
        edges.clear();
        rep1(i, n * m)
            g[i].clear();
        while (k--)
        {
            scanf("%d %d", &a, &b);
            land[a][b] = 1;
        }
        rep1(i, n)
        {
            rep1(j, m)
            {
                if (land[i][j] == 1)
                    continue;
                if (i > 1 && land[i - 1][j] == 0)
                    addEdge((i - 1) * m + j, (i - 2) * m + j);
                if (j > 1 && land[i][j - 1] == 0)
                    addEdge((i - 1) * m + j, (i - 1) * m + j - 1);
            }
        }
        printf("%d\n", hungarian());


    }

    return 0;
}