codevs 1922 骑士共存问题 网络流

题目链接

给一个n*n的棋盘， 上面有障碍物， 有障碍物的不能放东西。然后往上面放马， 马不能互相攻击， 问最多可以放多少个马。

按x+y的奇偶来划分， 如果两个格子可以互相攻击， 就连一条权值为1的边。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {-, -}, {, }, {, -}, {, -}, {, }, {-, -}, {-, } };
const int maxn = 2e5;
int n, a[][];
int q[maxn*], head[], dis[], s, t, num, m;
struct node
{
    int to, nextt, c;
    node(){}
    node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
}e[maxn*];
void init() {
    num = ;
    mem1(head);
}
void add(int u, int v, int c) {
    e[num] = node(v, head[u], c); head[u] = num++;
    e[num] = node(u, head[v], ); head[v] = num++;
}
int bfs() {
    mem(dis);
    dis[s] = ;
    int st = , ed = ;
    q[ed++] = s;
    while(st<ed) {
        int u = q[st++];
        for(int i = head[u]; ~i; i = e[i].nextt) {
            int v = e[i].to;
            if(!dis[v]&&e[i].c) {
                dis[v] = dis[u]+;
                if(v == t)
                    return ;
                q[ed++] = v;
            }
        }
    }
    return ;
}
int dfs(int u, int limit) {
    if(u == t) {
        return limit;
    }
    int cost = ;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(e[i].c&&dis[v] == dis[u]+) {
            int tmp = dfs(v, min(limit-cost, e[i].c));
            if(tmp>) {
                e[i].c -= tmp;
                e[i^].c += tmp;
                cost += tmp;
                if(cost == limit)
                    break;
            } else {
                dis[v] = -;
            }
        }
    }
    return cost;
}
int dinic() {
    int ans = ;
    while(bfs()) {
        ans += dfs(s, inf);
    }
    return ans;
}
int check(int x, int y)
{
    if(x >=  && y >=  && x < n && y < n && !a[x][y])
        return ;
    return ;
}
void solve()
{
    init();
    s = n*n, t = s+;
    for(int i = ; i < n; i++) {
        for(int j = ; j < n; j++) {
            if(a[i][j])
                continue;
            if((i+j)%) {
                add(s, i*n+j, );
                for(int k = ; k < ; k++) {
                    int x = i+dir[k][];
                    int y = j+dir[k][];
                    if(check(x, y)) {
                        add(i*n+j, x*n+y, );
                    }
                }
            } else {
                add(i*n+j, t, );
            }
        }
    }
    int ans = n*n-dinic()-m;
    cout<<ans<<endl;
}
int main()
{
    int x, y;
    while(~scanf("%d%d", &n, &m)) {
        mem(a);
        for(int i = ; i < m; i++) {
            scanf("%d%d", &x, &y);
            x--;
            y--;
            a[x][y] = ;
        }
        solve();
    }
    return ;
}