POJ - 3020 ? Antenna Placement 二分图最大匹配

http://poj.org/problem?id=3020

首先注意到，答案的最大值是‘*‘的个数，也就是相当于我每用一次那个技能，我只套一个‘*‘，是等价的。

所以，每结合一对**，则可以减少一次使用，所以就是找**的最大匹配数目。

对于每一个*，和它的上下左右连接一条边（如果是*才连）

那么，这个图是一个二分图，怎么找到左边集合S，右边集合T呢？

我的做法是染色一次，就可以。

这题应该不能贪心吧。

3 5

*****

o***o

o*o*o

其实也可以不分开S、T

跑一发最大匹配，然后匹配数 / 2即可。意思就是match[1] = 2，也可以match[2] = 1，但是两者是只算一个匹配。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 40 + 20;
char str[maxn][maxn];
int a[maxn][maxn];
struct Node {
    int u, v, tonext;
}e[maxn * maxn * 3];
int first[maxn * maxn * 3], num;
void addEdge(int u, int v) {
    ++num;
    e[num].u = u, e[num].v = v;
    e[num].tonext = first[u];
    first[u] = num;
}
int match[maxn * maxn * 3], book[maxn * maxn * 3], DFN;
bool dfs(int u) {
    for (int i = first[u]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (book[v] == DFN) continue;
        book[v] = DFN;
        if (match[v] == 0 || dfs(match[v])) {
            match[v] = u;
            return true;
        }
    }
    return false;
}
vector<int>vc;
int hungary() {
    memset(match, 0, sizeof match);
    int ans = 0;
    for (int i = 0; i < vc.size(); ++i) {
        ++DFN;
        if (dfs(vc[i])) ans++;
    }
    return ans;
}
int col[maxn * maxn * 3];
void ran(int cur, int which) {
    col[cur] = which;
    book[cur] = DFN;
    for (int i = first[cur]; i; i = e[i].tonext) {
        int v = e[i].v;
        if (book[v] == DFN) continue;
        ran(v, !which);
    }
}
void work() {
    memset(first, 0, sizeof first);
    num = 0;
    int to = 0;
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        scanf("%s", str[i] + 1);
    }
    memset(a, 0, sizeof a);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if (str[i][j] == ‘*‘) {
                a[i][j] = ++to;
            }
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            if (a[i][j + 1]) addEdge(a[i][j], a[i][j + 1]);
            if (a[i + 1][j]) addEdge(a[i][j], a[i + 1][j]);
            if (a[i][j - 1]) addEdge(a[i][j], a[i][j - 1]);
            if (a[i - 1][j]) addEdge(a[i][j], a[i - 1][j]);
        }
    }
    memset(col, -1, sizeof col);
    ++DFN;
    for (int i = 1; i <= to; ++i) {
        if (book[i] != DFN) ran(i, 0);
    }
    vc.clear();
    for (int i = 1; i <= to; ++i) {
        if (col[i] == 1) vc.push_back(i);
    }
    cout << to - hungary() << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    cin >> t;
    while (t--) work();
    return 0;
}