UVaLive 6525 Attacking rooks (二分图最大匹配)

时间：2023-03-10 03:03:21
UVaLive 6525 Attacking rooks (二分图最大匹配)

题意：给定一个 n * n的图，X是卒， . 是空位置，让你放尽量多的车，使得他们不互相攻击。

析：把每行连续的 . 看成X集体的一个点，同理也是这样，然后求一个最大匹配即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

//#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 100 + 10;

const int maxm = 3e5 + 10;

const ULL mod = 3;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

char s[maxn][maxn];

int a[maxn][maxn], b[maxn][maxn];

struct Edge{

  int to, next;

};

Edge edges[maxn*maxn];

int head[maxn*maxn], cnt;

void addEdge(int u, int v){

  edges[cnt].to = v;

  edges[cnt].next = head[u];

  head[u] = cnt++;

}

int match[maxn*maxn];

bool used[maxn*maxn];

bool dfs(int u){

  used[u] = 1;

  for(int i = head[u]; ~i; i = edges[i].next){

    int v = edges[i].to, w = match[v];

    if(w == -1 || !used[w] && dfs(w)){

      match[u] = v;

      match[v] = u;

      return true;

    }

  }

  return false;

}

int main(){

  while(scanf("%d", &n) == 1){

    int idx = 0;

    for(int i = 0; i < n; ++i){

      scanf("%s", s[i]);

      if(s[i][0] == '.')  a[i][0] = ++idx;

      for(int j = 1; j < n; ++j)

        if(s[i][j] == '.' && s[i][j] == s[i][j-1])  a[i][j] = a[i][j-1];

        else if(s[i][j] == '.')  a[i][j] = ++idx;

    }

    int row = idx;

    for(int j = 0; j < n; ++j){

      if(s[0][j] == '.')  b[0][j] = ++idx;

      for(int i = 1; i < n; ++i)

        if(s[i][j] == '.' && s[i][j] == s[i-1][j])  b[i][j] = b[i-1][j];

        else if(s[i][j] == '.') b[i][j] = ++idx;

    }

    cnt = 0;  ms(head, -1);

    FOR(i, 0, n)  FOR(j, 0, n) if(s[i][j] == '.'){

      addEdge(a[i][j], b[i][j]);

      addEdge(b[i][j], a[i][j]);

    }

    int ans = 0;  ms(match, -1);

    for(int i = 1; i <= row; ++i)  if(match[i] < 0){

      ms(used, 0);  if(dfs(i))  ++ans;

    }

    printf("%d\n", ans);

  }

  return 0;

}

　　

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