网络流24题

骑士共存问题

按马的轨迹连边，最后是一张二分图，求最大匹配。可以跑匈牙利直接求出来，也可以连源点和汇点跑Dinic。

我是学二分图的时候写的，当时写的是匈牙利。我当时码风好奇怪

#include <bits/stdc++.h>
#define ID(x, y) (x - 1) * n + y
#define New(p) p = &tmp[++ecnt]

struct Edge {int to; Edge *nxt;} tmp[1000005], *head[40005];

int vis[40005], match[40005], ecnt = -1, n, m, tim, ans, tot;
int dx[] = {-1, -2, 1, 2, -1, -2, 1, 2}, dy[]={-2, -1, -2, -1, 2, 1, 2, 1};
bool ban[205][205];

inline void Add(int f, int to) {Edge *p; New(p); p -> to = to, p -> nxt = head[f], head[f] = p;}

bool dfs(int x) {
    for (Edge *i = head[x]; i != NULL; i = i -> nxt) {
        if (vis[i -> to] != tim) {
            vis[i -> to] = tim;
            if (!match[i -> to] || dfs(match[i -> to])) {
                match[i -> to] = x;
                return 1;
            }
        }
    }
    return 0;
}

signed main() {
    scanf("%d%d", &n, &m);
    for (int i = 1, x, y; i <= m; i++) {
        scanf("%d%d", &x, &y);
        ban[x][y] = 1;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if(ban[i][j] || ((i + j) & 1)) continue;
            for (int k = 0; k < 8; k++) {
                int x = i + dx[k], y = j + dy[k];
                if (ban[x][y] || x < 1 || x > n || y < 1 || y > n) continue;
                Add(ID(i, j), ID(x, y));
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (!ban[i][j]) {
                ++tim;
                ++tot;
                if (dfs(ID(i, j))) ++ans;
            }
        }
    }
    printf("%d
", tot - ans);
    return 0;
}

飞行员配对方案问题

按给定关系建图，仍是二分图，求最大匹配。

依旧写的是匈牙利。我当时没看见SPJ，写了一堆奇怪的东西。。。。

#include <bits/stdc++.h>
using namespace std;
#define N 1005

namespace Gekoo {
    struct Edge {
        int to, nxt;
    }e[N];

    pair<int, int> p[N];

    int m, n, head[N], ecnt, match[N], ans, pcnt;
    bool vis[N];

    void AddEdge(int f, int to) {
        e[++ecnt].to = to;
        e[ecnt].nxt = head[f];
        head[f] = ecnt;
    }

    bool dfs(int u) {
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to;
            if (!vis[v]) {
                vis[v] = 1;
                if (!match[v] || dfs(match[v])) {
                    match[v] = u;
                    return 1;
                }
            }
        }
        return 0;
    }

    void QAQ() {
        scanf("%d%d", &m, &n);
        int i, j;
        while (1) {
            scanf("%d%d", &i, &j);
            if (i == -1 && j == -1) break;
            else AddEdge(i, j);
        }
        for (int i = 1; i <= m; i++) {
            memset(vis, 0, sizeof(vis));
            if(dfs(i)) ans++;
        }
        if (ans == 0) {
            puts("No Solution!");
            return ;
        } else {
            printf("%d
", ans);
            for (int i = m + 1; i <= n; i++) {
                if (match[i]) {
                    p[++pcnt] = make_pair(match[i], i);
                }
            }
            sort(p + 1, p + 1 + pcnt);
            for (int i = 1; i <= pcnt; i++) {
                printf("%d %d
", p[i].first, p[i].second);
            }
            return ;
        }
    }
}

signed main() {
    Gekoo::QAQ();
    return 0;
}

方格取数问题

把图按格子奇偶性黑白染色，源点连黑色，白色连汇点，黑色向相邻白点连，求出最小割，用数字和减去。

建出来的仍是二分图，这个过程其实就是在求二分图的最大独立集。

#include <bits/stdc++.h>
#define id(x, y) ((x - 1) * m + y) 

const int N = 500005, INF = 0x3f3f3f3f;
const int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
int n, m, sum, ecnt = 1, S, T;
int matrix[105][105], head[N], dep[N];
struct Edge {
    int to, nxt, val;
} e[N << 5];

inline void addEdge(int f, int to, int val) {
    e[++ecnt] = {to, head[f], val}, head[f] = ecnt;
    e[++ecnt] = {f, head[to], 0}, head[to] = ecnt;
}

bool bfs() {
    memset(dep, 0, sizeof(dep));
    dep[S] = 1;
    std::queue<int> q;
    q.push(S);
    while (!q.empty()) {
        int x = q.front();
        q.pop();
        for (int i = head[x], y = e[i].to; i; i = e[i].nxt, y = e[i].to) {
            if (!dep[y] && e[i].val) {
                dep[y] = dep[x] + 1;
                if (y == T) return true;
                q.push(y);
            }
        }
    }
    return false;
}

int dfs(int x, int flow) {
    if (x == T) return flow;
    int las = flow, q;
    for (int i = head[x], y = e[i].to; i; i = e[i].nxt, y = e[i].to) {
        if (dep[y] == dep[x] + 1 && e[i].val && las) {
            q = dfs(y, std::min(e[i].val, las));
            if (!q) {dep[y] = 0; continue;}
            las -= q, e[i].val -= q, e[i^1].val += q;
        }
    }
    return flow - las;
}

int dinic() {
    int maxflow = 0;
    while (bfs()) maxflow += dfs(S, INF);
    return maxflow;
}

signed main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%d", &matrix[i][j]);
            sum += matrix[i][j];
        }
    }
    S = n * m  + 1, T = S + 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if ((i + j) & 1) {
                addEdge(S, id(i, j), matrix[i][j]);
                for (int k = 0; k <= 3; k++) {
                    int xx = i + dx[k], yy = j + dy[k];
                    if (xx < 1 || yy < 1 || xx > n || yy > m) continue;
                    addEdge(id(i, j), id(xx, yy), INF);
                }
            } else {
                addEdge(id(i, j), T, matrix[i][j]);
            }
        }
    }
    printf("%d
", sum - dinic());
    return 0;
}