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('\n')
#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;
}
posted @ 2017-12-27 16:28  胡小兔  阅读(243)  评论(0编辑  收藏  举报