[ZJOI2009]狼和羊的故事【最小割】

題目鏈接 P2598 [ZJOI2009]狼和羊的故事


  要讓羊和狼都區別開來,需要的最小的割是多少?每隻羊向四周有4個可能的方向,每隻狼也是同樣的,所以每個動物向周圍可以跑出4個流,我們要建立柵欄,可以考慮成切斷的“割”,那麼最少需要多少的割才能使得羊狼無法接觸呢?於是我們對羊連接在源點上,狼連接在匯點上,羊抵達不了狼,那麼就是說明是其中的一種割的方案了,最小的割就是最大流了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e4 + 7, maxM = 1e5 + 7;
const int dir[4][2] =
{
    -1, 0,
    0, -1,
    0, 1,
    1, 0
};
int N, M, mp[105][105];
inline int _ID(int x, int y) { return (x - 1) * M + y; }
inline bool In_Map(int x, int y) { return x >= 1 && y >= 1 && x <= N && y <= M; }
int S, T, head[maxN], cnt, cur[maxN];
struct Eddge
{
    int nex, to; ll flow;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), flow(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, 0); }
struct Max_Flow
{
    int gap[maxN], d[maxN], que[maxN], ql, qr, node;
    inline void init()
    {
        for(int i=0; i<=node + 1; i++)
        {
            gap[i] = d[i] = 0;
            cur[i] = head[i];
        }
        ++gap[d[T] = 1];
        que[ql = qr = 1] = T;
        while(ql <= qr)
        {
            int x = que[ql ++];
            for(int i=head[x], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(!d[v]) { ++gap[d[v] = d[x] + 1]; que[++qr] = v; }
            }
        }
    }
    inline ll aug(int x, ll FLOW)
    {
        if(x == T) return FLOW;
        int flow = 0;
        for(int &i=cur[x], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(d[x] == d[v] + 1)
            {
                ll tmp = aug(v, min(FLOW, edge[i].flow));
                flow += tmp; FLOW -= tmp; edge[i].flow -= tmp; edge[i ^ 1].flow += tmp;
                if(!FLOW) return flow;
            }
        }
        if(!(--gap[d[x]])) d[S] = node + 1;
        ++gap[++d[x]]; cur[x] = head[x];
        return flow;
    }
    inline ll max_flow()
    {
        init();
        ll ret = aug(S, INF);
        while(d[S] <= node) ret += aug(S, INF);
        return ret;
    }
} mf;

inline void init()
{
    cnt = 0; S = 0; T = N * M + 1; mf.node = T + 1;
    for(int i=0; i<=mf.node; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d", &N, &M);
    init();
    for(int i=1; i<=N; i++)
    {
        for(int j=1; j<=M; j++)
        {
            scanf("%d", &mp[i][j]);
        }
    }
    for(int x=1, xx, yy; x<=N; x++)
    {
        for(int y=1; y<=M; y++)
        {
            if(mp[x][y] == 1) _add(S, _ID(x, y), 4);
            else if(mp[x][y] == 2) _add(_ID(x, y), T, 4);
            for(int i=0; i<4; i++)
            {
                xx = x + dir[i][0]; yy = y + dir[i][1];
                if(!In_Map(xx, yy)) continue;
                _add(_ID(x, y), _ID(xx, yy), 1);
            }
        }
    }
    printf("%lld\n", mf.max_flow());
    return 0;
}
/*
3 3
1 0 1
2 0 2
1 1 1
ans:6
*/

 

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