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Luogu P3159 [CQOI2012]交换棋子

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本文字数： 3k 阅读时长 ≈ 3 分钟

题目描述

https://www.luogu.com.cn/problem/P3159

Solution

我们发现只需要将黑色棋子移动到对应位置即可

对于一条交换路径，终点和起点都只交换一次，其它位置交换两次

我们考虑拆点来限制流量

对于非起点和非终点的点，流量为 $\frac{c[i][j]}{2} $

对于起点或者终点的点，由于这些点只会在一次交换中作为起点或者终点，所以它的流量应该为 $\frac{c[i][j]-1}{2}+1$，实际上就是 $\frac{c[i][j]+1}{2}$

如果起始状态和终止状态都是黑点的话，我们直接看做白点就行

建图：

$s$ 连 起始状态是黑点的点，容量为 $1$，费用为 $0$

终止状态时黑点的点 连 $t$，容量为 $1$，费用为 $0$

每个点拆点，容量与上文分析相同，费用为 $0$

每个点的出点 连 周围八连通的点的入点，容量为 $\infty$，费用为 $1$

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#include <iostream>
#include <cstdio>
#include <queue>
#define maxn 810
#define maxm 21
#define INF 1000000000
using namespace std;

int dx[] = { 0, 0, 0, 1, 1, 1, -1, -1, -1 };
int dy[] = { 1, -1, 0, 1, -1, 0, 1, -1, 0 }; 

int n, m, s, t;

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

inline int id(int x, int y) { return (x - 1) * m + y; } 

struct Edge {
    int to, next, w, fi; 
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w, int fi) {
    e[c1].to = v; e[c1].w = w; e[c1].fi = fi;
    e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v,  int w, int fi) {
    add_edge(u, v, w, fi); add_edge(v, u, 0, -fi);
}

int dis[maxn], la[maxn], fl[maxn], pre[maxn]; bool vis[maxn];  
bool spfa() {
    fill(dis, dis + maxn, INF); dis[s] = 0;
    fill(vis, vis + maxn, 0); vis[s] = 1; 
    queue<int> Q; Q.push(s); pre[t] = -1; fl[s] = INF;
    while (!Q.empty()) {
        int u = Q.front(); Q.pop(); vis[u] = 0; 
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to, w = e[i].w, fi = e[i].fi;
            if (w > 0 && dis[v] > dis[u] + fi) {
                dis[v] = dis[u] + fi; pre[v] = u; 
                fl[v] = min(fl[u], w); la[v] = i;
                if (vis[v]) continue; vis[v] = 1;
                Q.push(v);
            }
        }
    }
    return ~pre[t]; 
}

int mcmf(int _mf) {
    int mf = 0, mc = 0;
    while (spfa()) {
        mf += fl[t];
        mc += dis[t] * fl[t]; 
        int now = t;
        while (now != s) {
            e[la[now]].w -= fl[t];
            e[la[now] ^ 1].w += fl[t];
            now = pre[now];
        }
    }
    return mf == _mf ? mc : -1;
} 

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m; int s1 = 0, s2 = 0, s3 = 0;
    s = 0; t = 2 * n * m + 1; 
    for (int i = 1; i <= n; ++i) {
        char s[maxm]; cin >> s + 1;
        for (int j = 1; j <= m; ++j) a[i][j] = s[j] - '0'; 
    }
    for (int i = 1; i <= n; ++i) {
        char s[maxm]; cin >> s + 1;
        for (int j = 1; j <= m; ++j) b[i][j] = s[j] - '0';
    }
    for (int i = 1; i <= n; ++i) {
        char s[maxm]; cin >> s + 1; 
        for (int j = 1; j <= m; ++j) c[i][j] = s[j] - '0';
    } 
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) { 
            s1 += a[i][j], s2 += b[i][j], s3 += a[i][j] & b[i][j];
            if (a[i][j] && b[i][j]) a[i][j] = b[i][j] = 0;
            else if (a[i][j] || b[i][j]) ++c[i][j];
        } 
    if (s1 != s2) return cout << "-1\n", 0; s1 -= s3; 
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) {
            Add_edge(id(i, j), id(i, j) + n * m, c[i][j] / 2, 0); 
            if (a[i][j]) Add_edge(s, id(i, j), 1, 0);
            if (b[i][j]) Add_edge(id(i, j) + n * m, t, 1, 0); 
            for (int k = 0; k < 9; ++k) {
                int x = i + dx[k], y = j + dy[k];
                if (x < 1 || x > n || y < 1 || y > m) continue;
                Add_edge(id(i, j) + n * m, id(x, y), INF, 1); 
            }
        }
    cout << mcmf(s1) << "\n"; 
    return 0;
}