• bzoj3144


    最小割

    问题就在于如何限制<=D

    http://blog.csdn.net/zarxdy34/article/details/45272055

    只要仔细看看那个图就懂了

    #include<bits/stdc++.h>
    using namespace std;
    const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
    const int N = 41 * 41 * 41, inf = 1e9;
    int rd()
    {
        int x = 0, f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
        return x * f;
    }
    int n, m, k, D, source, sink, cnt = 1, tot;
    int a[N], d[N], head[N], iter[N], id[41][41][41];
    struct edge {
        int nxt, to, f;
    } e[N * 5];
    bool bfs()
    {
        queue<int> q;
        memset(d, -1, sizeof(d));
        d[source] = 0;
        q.push(source);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            for(int i = head[u]; i; i = e[i].nxt) if(d[e[i].to] == -1 && e[i].f) 
            {
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
        return d[sink] != -1;
    }
    int dfs(int u, int delta)
    {
        if(u == sink) return delta;
        int ret = 0;
        for(int &i = iter[u]; i && delta; i = e[i].nxt) if(d[e[i].to] == d[u] + 1 && e[i].f)
        {
            int x = dfs(e[i].to, min(delta, e[i].f));
            ret += x;
            delta -= x;
            e[i].f -= x;
            e[i ^ 1].f += x;    
        }
        return ret;
    }
    int dinic() 
    {
        int ret = 0;
        while(bfs())
        {
            for(int i = source; i <= sink; ++i) iter[i] = head[i];
            ret += dfs(source, inf);
        }
        return ret;
    }
    void link(int u, int v, int f)
    {
        e[++cnt].nxt = head[u];
        head[u] = cnt;
        e[cnt].to = v;
        e[cnt].f = f;
    }
    void insert(int u, int v, int f)
    {
        link(u, v, f);
        link(v, u, 0);
    }
    int main()
    {
        scanf("%d%d%d%d", &n, &m, &k, &D);
        sink = n * m * (k + 2) + 1;
        for(int x = 1; x <= n; ++x)
            for(int y = 1; y <= m; ++y)
                for(int z = 1; z <= k + 1; ++z)
                    id[x][y][z] = ++tot;
        sink = ++tot;
        for(int z = 1; z <= k; ++z) 
            for(int x = 1; x <= n; ++x)
                for(int y = 1; y <= m; ++y) 
                {
                    int v;
                    scanf("%d", &v);
                    insert(id[x][y][z], id[x][y][z + 1], v);
                    for(int i = 0; i < 4; ++i) 
                    {
                        int xx = x + dx[i], yy = y + dy[i];
                        if(xx > 0 && xx <= n && yy > 0 && yy <= m) 
                            if(z - 1 - D > 0) insert(id[x][y][z], id[xx][yy][z - D], inf);
                    }
                }
        for(int x = 1; x <= n; ++x)
            for(int y = 1; y <= m; ++y) 
            {
                insert(source, id[x][y][1], inf);
                insert(id[x][y][k + 1], sink, inf);
            }
        printf("%d
    ", dinic());
        return 0;
    }
    
    View Code
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  • 原文地址:https://www.cnblogs.com/19992147orz/p/7978469.html
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