[ZOJ1002] Fire Net

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题目链接：

走这里

题目分析：

联想到Asteroids这道题，将行和列分别作为二分图的两边的点，把点本身作为二分图的边去处理
那么这个题的限制条件打不穿的墙怎么处理呢，发现这样的话
\[ ...X.. \]
其实左边那段和右边那段在横着考虑的时候是互不影响的，不妨对行和列重新编号

用一下某题解的剪枝讲解的图，样例第一个大概就是这个样子
然后对于每个空地的行编号和列编号连一条边，当空地已经被控制/放满了就是最终答案，跑最大匹配即可

代码：

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
#define N (800 + 10)
using namespace std;
inline int read() {
    int cnt = 0, f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
    while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
    return cnt * f;
}
int n, cnt = 0, res = 0, l, r, ans;
char s[N][N];
int mapp[N][N], mapp2[N][N];
bool vis[N];
int nxt[N * 100], first[N * 100], to[N * 100], tot;
int match[N];
void add(int x, int y) {nxt[++tot] = first[x], first[x] = tot, to[tot] = y;}
void build_map() {
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= n; ++j) {
            if (s[i][j] == 'X') {
                mapp[i][j] = 0;
                continue;
            } else {
                if (s[i][j] == '.' && s[i][j - 1] == 'X') ++cnt;
                mapp[i][j] = cnt;
            }
        }
    l = cnt, cnt = 0;
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= n; ++j) {
            if (s[j][i] == 'X') {
                mapp2[j][i] = 0;
                continue;
            } else {
                if (s[j][i] == '.' && s[j - 1][i] == 'X') ++cnt;
                mapp2[j][i] = cnt;
            }
        }
    r = cnt, cnt = 0;
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= n; ++j) {
            if (!mapp[i][j]) continue;
            add(mapp[i][j], mapp2[i][j]);
        }
}

bool find (int u) {
    for (register int i = first[u]; i; i = nxt[i]) {
        int v = to[i];
        if (vis[v]) continue;
        else {
            vis[v] = true;
            if (match[v] == -1 || find(match[v])) {
                match[v] = u;
                return 1;
            }
        }
    }
    return 0;
}
int hungary() {
    ans = 0;
    for (register int i = 1; i <= l; ++i) {
        memset(vis, 0, sizeof(vis));
        ans += find(i);
    }
    return ans;
}
int main() {
//  freopen("1.in", "r", stdin);
  while (1) {
    n = read();
    if (!n) break;
    memset(first, 0, sizeof (first));
    memset(to, 0, sizeof(to));
    memset(nxt, 0, sizeof(nxt));
    tot = 0, l = 0, r = 0;
    for (register int i = 0; i <= n + 1; ++i)
        for (register int j = 0; j <= n + 1; ++j) s[i][j] = 'X';
    for (register int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
    for (register int i = 0; i <= n + 1; ++i) s[i][n + 1] = 'X';
    build_map();
    for (register int i = 1; i <= r; ++i) match[i] = -1;
    res = hungary();
    printf("%d\n", res);
  }
    return 0;
}

[ZOJ1002] Fire Net

标签：href   控制   getc   void   main   技术   tin   eof   info   

原文地址：https://www.cnblogs.com/kma093/p/11518284.html