[网络流24题]P3355 骑士共存

最小割经典思想+用计算机系统的内存访问卡常
https://www.luogu.com.cn/problem/P3355

思路

考虑首先将所有的方格都选上，再去掉不能选的部分。

我们将所有互斥的方格之间连边，当图连通时则表示当前解中仍然存在不合法的放法，因此很自然地想到当图被割开时合法，由于要最大化答案，因此求最小割。
不难发现棋盘格的黑白两色形成一个二分图，很好建图，最大匹配/最大流即可，这题的卡常姿势很有趣，计系实验过后第一次在acm里用这个2333。

代码

来看一看下面两个存图时的访问顺序

int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};

和

int dx[8] = {1, 2, 2, 1, -1, -2, -2, -1};
int dy[8] = {-2, -1, 1, 2, -2, -1, 1, 2};

上面的写法本地运行4000+，而下面的只需要 700+，在计系中学习过连续内存访问可以提高程序运行效率，这可以快乐地从TLE变成AC！表示学到了，下次可以 next_permutation 一组最快的提交2333

二分图卡常代码

#include <bits/stdc++.h>
using namespace std;

bool vis[40010];
int link[40010];
int n, m, ans;
vector<int> mp[40010];
bool f[205][205];
/*
int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};*/

const int dx[8] = {1, 2, 2, 1, -1, -2, -2, -1};
const int dy[8] = {-2, -1, 1, 2, -2, -1, 1, 2};

int dfs(int x) {
    for (int i = 0; i < mp[x].size(); i++) {
        int v = mp[x][i];
        if (vis[v] == 0) {
            vis[v] = 1;
            if (link[v] == 0 || dfs(link[v]) == 1) {
                link[v] = x;
                return 1;
            }
        }
    }
    return 0;
}

int id(int x, int y) { return (x - 1) * n + y; }
clock_t st, ed;
// int node[N], tot;
int main() {
    st = clock();
    freopen("C:\\Users\\lenovo\\Downloads\\P3355_3.in", "r", stdin);
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int x, y;
        cin >> x >> y;
        f[x][y] = 1;
    }
    // cout << "aaa";
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if ((i + j) % 2 || f[i][j]) continue;
            for (int k = 0; k < 8; k++) {
                int nx = i + dx[k], ny = j + dy[k];
                if (nx > 0 && ny > 0 && nx <= n && ny <= n && !f[nx][ny])
                    mp[id(i, j)].push_back(id(nx, ny));
            }
        }
    }

    for (int i = 1; i <= n * n; i++) {
        memset(vis, 0, sizeof(vis));
        if (dfs(i)) ans++;
    }

    cout << n * n - m - ans << endl;
    ed = clock();
    cout << (ed - st) << endl;
}

正经Dinic

点击查看代码

#include <bits/stdc++.h>
#define endl '\n'
#define IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0)
#define P pair<int, int>
#define endl '\n'
using namespace std;

typedef long long ll;
const int maxn = 200 * 200 + 10;
const int N = 200 * 200 + 10;
const int M = N << 5;
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
ll dis[maxn];
bool vis[maxn];
struct node {
    int v, w, to;
} edge[M * 2];
int pre[N], cnt, dep[N];
int S, T, head[N], sum;
int n, m, q[N], cur[N];
void add(int u, int v, int w) {
    // cout << u << " " << v << endl;
    edge[cnt] = {v, w, head[u]};
    head[u] = cnt++;
    edge[cnt] = {u, 0, head[v]};
    head[v] = cnt++;
}
bool bfs() {
    for (int i = 0; i <= T; i++) dep[i] = 0;
    dep[S] = 1;
    int l = 0, r = 1;
    q[r] = S;
    while (l < r) {
        int u = q[++l];
        for (int i = head[u]; i != -1; i = edge[i].to) {
            int v = edge[i].v;
            if (!dep[v] && edge[i].w) dep[v] = dep[u] + 1, q[++r] = v;
        }
    }
    return dep[T];
}
int dfs(int u, int mi) {
    int res = 0;
    if (mi == 0 || u == T) return mi;
    for (int &i = cur[u]; i != -1; i = edge[i].to) {
        int v = edge[i].v;
        if (dep[u] + 1 == dep[v] && edge[i].w) {
            int minn = dfs(v, min(mi - res, edge[i].w));
            edge[i].w -= minn;
            edge[i ^ 1].w += minn;
            res += minn;
            if (res == mi) return res;
        }
    }
    if (res == 0) dep[u] = 0;
    return res;
}
int dinic() {
    ll res = 0;
    while (bfs()) {
        memcpy(cur, head, sizeof(head));
        //        cout<<res<<endl;
        res += dfs(S, INF);
    }
    return res;
}
int id(int x, int y) { return (n * (x - 1) + y); }
int mp[205][205], f;
int dx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[] = {1, 2, 2, 1, -1, -2, -2, -1};
int main() {
    freopen("C:\\Users\\lenovo\\Downloads\\P3355_3.in", "r", stdin);
    memset(head, -1, sizeof head);
    cin >> n >> m;
    int ans = n * n;
    S = ans + 1, T = S + 1;
    for (int i = 0, x, y; i < m; i++) {
        cin >> x >> y;
        mp[x][y] = 1;
        ans--;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (mp[i][j]) continue;
            int now = id(i, j);
            if ((i+j) & 1) {
                if (!mp[i][j]) add(S, now, 1);
            } else {
                if (!mp[i][j]) add(now, T, 1);
                continue;
            }

            for (int k = 0; k < 8; k++) {
                int nx = i + dx[k], ny = j + dy[k];
                if (nx < 1 || nx > n || ny < 1 || ny > n || mp[nx][ny])
                    continue;

                int to = id(nx, ny);
                add(now, to, INF);

                // printf("(%d,%d) :%d\n", i, j, id(i, j));
                // printf("(%d,%d) :%d\n", nx, ny, id(nx, ny));
                // add(now, id(nx, ny), INF);
            }
        }
    }
    int flow = dinic();
    // cout << flow << endl;
    ans -= flow;
    cout << ans;
    return 0;
}