[TJOI2007]脫險【最大流】

題目鏈接

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  首先，每個時刻，每個點最多經過一個人，那麼來說，我們需要對於每個時刻的每個點進行拆點，限制流量。

  然後，因爲圖實際上只有10*10，因爲最外圍是包圍圈，然後時間上其實是上限50，所以我們不妨對於時間來建點，表示時間爲Tim時候的座標爲（x，y）的點，那麼，根據每一秒移動一步，初始爲0，我們向四個方向都是鏈接上下一秒時刻對應的點值。

  接下去，就是最大流的部分了。

#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 0x3f3f3f3f
#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)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int dir[4][2] =
{
    -1, 0,
    0, -1,
    0, 1,
    1, 0
};
const int maxN = 2e4 + 7, maxM = 2e5 + 7;
int R, C, Tm;
inline bool In_Map(int x, int y) { return x >= 0 && y >= 0 && x < R && y < C; }
int head[maxN], cnt;
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, S, T, cur[maxN], 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;
int tot, ID[15][15];
char mp[15][15];
int main()
{
    scanf("%d%d%d", &R, &C, &Tm);
    for(int i=0; i<R; i++)
    {
        scanf("%s", mp[i]);
        for(int j=0; j<C; j++)
        {
            if(mp[i][j] == '*') continue;
            ID[i][j] = ++tot;
        }
    }
    mf.S = 0; mf.T = 2 * (Tm + 1) * tot + 1; mf.node = mf.T + 1;
    for(int i=0; i<=mf.node + 1; i++) head[i] = -1;
    for(int i=0; i<R; i++)
    {
        for(int j=0; j<C; j++)
        {
            if(mp[i][j] == 'P')
            {
                _add(mf.S, ID[i][j], 1);
            }
            else if(mp[i][j] == 'O')
            {
                _add(ID[i][j], mf.T, INF);
            }
        }
    }
    for(int tim = 0; tim < Tm; tim++)
    {
        for(int i=0, x, y; i<R; i++)
        {
            for(int j=0; j<C; j++)
            {
                if(mp[i][j] == '*' || mp[i][j] == 'O') continue;
                _add(tim * tot * 2 + ID[i][j], tim * tot * 2 + tot + ID[i][j], 1);  //capa limited
                _add(tim * tot * 2 + tot + ID[i][j], (tim + 1) * tot * 2 + ID[i][j], 1);    //stay here
                for(int k=0; k<4; k++)
                {
                    x = i + dir[k][0]; y = j + dir[k][1];
                    if(!In_Map(x, y) || mp[x][y] == '*') continue;
                    if(mp[x][y] == 'O')
                    {
                        _add(tim * tot * 2 + tot + ID[i][j], ID[x][y], 1);  //move to next position
                    }
                    else _add(tim * tot * 2 + tot + ID[i][j], (tim + 1) * tot * 2 + ID[x][y], 1);
                }
            }
        }
    }
    printf("%lld\n", mf.max_flow());
    return 0;
}