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  • BZOJ 1412 [ZJOI2009]狼和羊的故事 | 网络流

    显然是个最小割嘛!

    一开始我是这么建图的:

    1. 源点向狼连INF
    2. 羊向汇点连INF
    3. 每两个相邻格子间连双向边,边权为1

    然后T成狗

    后来我是这么建图的:

    1. 源点向狼连INF
    2. 羊向汇点连INF
    3. 狼和空地向相邻的非狼节点连1

    然后跑得 跟HK Journalist 跟中午抢饭的国大班同学们一样快

    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    using namespace std;
    typedef long long ll;
    #define enter putchar('
    ')
    #define space putchar(' ')
    template <class T>
    void read(T &x){
        char c;
        bool op = 0;
        while(c = getchar(), c > '9' || c < '0')
    	if(c == '-') op = 1;
        x = c - '0';
        while(c = getchar(), c >= '0' && c <= '9')
    	x = x * 10 + c - '0';
        if(op) x = -x;
    }
    template <class T>
    void write(T x){
        if(x < 0) putchar('-'), x = -x;
        if(x >= 10) write(x / 10);
        putchar('0' + x % 10);
    }
    
    const int W = 105, N = 100005, M = 1000005, INF = 0x3f3f3f3f;
    const int dx[] = {0, 0, 1, -1};
    const int dy[] = {1, -1, 0, 0};
    int n, m, mp[W][W], src, des;
    int ecnt = 1, adj[N], cur[N], dis[N], go[M], nxt[M], cap[M];
    #define id(x, y) (((x) - 1) * m + (y))
    
    void ADD(int u, int v, int _cap){
        go[++ecnt] = v;
        nxt[ecnt] = adj[u];
        adj[u] = ecnt;
        cap[ecnt] = _cap;
    }
    void add(int u, int v, int _cap){
        ADD(u, v, _cap);
        ADD(v, u, 0);
    }
    bool bfs(){
        static int que[N], qr;
        for(int i = 1; i <= des; i++)
            dis[i] = -1, cur[i] = adj[i];
        dis[src] = 0, que[qr = 1] = src;
        for(int ql = 1; ql <= qr; ql++){
            int u = que[ql];
            for(int e = adj[u], v; e; e = nxt[e])
                if(cap[e] && dis[v = go[e]] == -1){
                    dis[v] = dis[u] + 1, que[++qr] = v;
                    if(v == des) return 1;
                }
        }
        return 0;
    }
    int dfs(int u, int flow){
        if(u == des) return flow;
        int ret = 0, delta;
        for(int &e = cur[u], v; e; e = nxt[e])
            if(cap[e] && dis[v = go[e]] == dis[u] + 1){
                delta = dfs(v, min(flow - ret, cap[e]));
                if(delta){
                    cap[e] -= delta;
                    cap[e ^ 1] += delta;
                    ret += delta;
                    if(ret == delta) return ret;
                }
            }
        dis[u] = -1;
        return ret;
    }
    int maxflow(){
        int ret = 0;
        while(bfs()){
            int flow;
            do{
                ret += (flow = dfs(src, INF));
            }while(flow);
        }
        return ret;
    }
    bool legal(int i, int j){
        return i > 0 && j > 0 && i <= n && j <= m && mp[i][j] != 1;
    }
    
    int main(){
    
        read(n), read(m), src = n * m + 1, des = src + 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                read(mp[i][j]);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++){
                if(mp[i][j] == 1) add(src, id(i, j), INF);
                else if(mp[i][j] == 2) add(id(i, j), des, INF);
                if(mp[i][j] != 2)
                    for(int k = 0, x, y; k < 4; k++)
                        if(legal(x = i + dx[k], y = j + dy[k]))
                            add(id(i, j), id(x, y), 1);
            }
        write(maxflow()), enter;
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/RabbitHu/p/BZOJ1412.html
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