poj 2226 Muddy Fields（水二分图）-优快云博客

本文介绍了一种解决农场泥泞区域覆盖问题的算法，通过使用Dinic算法找到最少数量的木板来覆盖所有泥泞区域，同时确保木板不覆盖草地。算法将连续的泥泞区域缩点并构建二分图进行匹配。

Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don't want to get their hooves dirty while they eat.

To prevent those muddy hooves, Farmer John will place a number of wooden boards over the muddy parts of the cows' field. Each of the boards is 1 unit wide, and can be any length long. Each board must be aligned parallel to one of the sides of the field.

Farmer John wishes to minimize the number of boards needed to cover the muddy spots, some of which might require more than one board to cover. The boards may not cover any grass and deprive the cows of grazing area but they can overlap each other.

Compute the minimum number of boards FJ requires to cover all the mud in the field.

Input

* Line 1: Two space-separated integers: R and C

* Lines 2..R+1: Each line contains a string of C characters, with '*' representing a muddy patch, and '.' representing a grassy patch. No spaces are present.

Output

* Line 1: A single integer representing the number of boards FJ needs.

Sample Input

4 4
*.*.
.***
***.
..*.

Sample Output

Hint

OUTPUT DETAILS:

Boards 1, 2, 3 and 4 are placed as follows:
1.2.
.333
444.
..2.
Board 2 overlaps boards 3 and 4.

虽然想到了做法但没想到和这题的联系细想才明白过来

一个提示

只要涉及这种类型的题一套Dinic就完事了、

每一行连续的点缩为一个点放在左边

每一列连续的点缩为一个点放在右边

匹配就好了

这里用的Dinic 匈牙利和hc也行

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 10000, INF = 0x7fffffff;
int n, m;

char str[55][55];
int nu1[55][55], nu2[55][55];

int vis[maxn], d[maxn], nex[maxn], head[maxn], cnt, cur[maxn];
int s, t;

struct node
{
    int u, v, c;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    mem(d, 0);
    queue<int> Q;
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = nex[i])
        {
            int v = Node[i].v;
            if(!d[v] && Node[i].c > 0)
            {
                d[v] = d[u] + 1;
                Q.push(v);
                if(v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(d[v] == d[u] + 1 && Node[i].c > 0)
        {
            int V = dfs(v, min(cap, Node[i].c));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            cap -= V;
            ret += V;
            if(cap == 0) break;
        }
    }
    return ret;
}

int Dinic()
{
    int ret = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof(head));
        ret += dfs(s, INF);
    }
    return ret;
}


int main()
{
    cnt = 0;
    mem(head, -1);
    rd(n), rd(m);
    s = 0, t = n * m + 1;
    rep(i, 0, n)
        scanf("%s", str[i]);
    int ans = 0;
    bool flag = 0;
    rep(i, 0, n)
    {
        //int j = 0;
        rep(j, 0, m)
        {
            if(str[i][j] == '*')
            {
                if(j == 0 || str[i][j] != str[i][j - 1])
                    ans++;

                nu1[i][j] = ans;
            }
        }
    }

    rap(i, 1, ans)
        add(s, i, 1);
    int ss = ans + 1;
    rep(j, 0, m)
    {
        rep(i, 0, n)
        {
            if(str[i][j] == '*')
            {

                if(i == 0 || str[i][j] != str[i - 1][j])
                    ans++;
                nu2[i][j] = ans;
            }
        }
    }
    rap(i, ss, ans)
        add(i, t, 1);
    rep(i, 0, n)
    {

        rep(j, 0, m)
        {
            if(str[i][j] == '*')
                add(nu1[i][j], nu2[i][j], 1);
        }
    }
    pd(Dinic());


    return 0;
}