hdu 2732 Leapin' Lizards 拆点最大流 isap

题意：n*m的矩阵有一些蜥蜴，每只蜥蜴最远能跳d（0~3）个单位，而每次起跳蜥蜴的站台会少一格血，最开始的时候蜥蜴保证站在有站台（站台初始血量为0~3）的地方。问最多有多少只蜥蜴能跳出矩阵。

建超源超汇，超源连有蜥蜴的地方，容量为1。站台之间相互可达的连边，容量为inf，跳出矩阵的点与超汇连边，容量也为inf。这时候将所有站台的点拆成入点和出点，入点到出点连边，容量就是这个站台的血量（理解成能通过这个站台的最大容量），那么站台之间连inf容量的边就很好理解了：站台之间可以无限次跳，但是某站台被经过的次数只能是它的血量。最后跑一发最大流即可。注意输出时候的单数复数

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define asd puts("sdfsdfsdfsdfsdfsdf");
typedef long long ll;
typedef pair<ll, int> pl;
const int inf = 0x3f3f3f3f;
const int N = 6400;
const int X = 45;
const int st = 0, ed = (N << 1) - 10;

struct node{
    int w, v, nxt;
}e[N<<3];

char a[X][X], b[X][X];
int n, m, cnt, k;
int dep[N<<1];
int gap[N<<1];
int head[N<<1];
int cur[N<<1];
int s[N<<1], top;
int tid[X][X];
queue <int> q;

void init()
{
    int cnt = 0;
    memset( head, -1, sizeof( head ) );
    memset( tid, 0, sizeof( tid ) );
}

void add( int u, int v, int w )
{
    e[cnt].w = w;
    e[cnt].v = v;
    e[cnt].nxt = head[u];
    head[u] = cnt++;

    e[cnt].w = 0;
    e[cnt].v = u;
    e[cnt].nxt = head[v];
    head[v] = cnt++;
}

int dis( int x, int y, int xx, int yy )
{
    return abs( x - xx ) + abs( y - yy );
}

void rev_bfs( )
{
    memset( gap, 0, sizeof( gap ) );
    memset( dep, -1, sizeof( dep ) );
    while( !q.empty() ) q.pop();
    q.push( ed );
    dep[ed] = 0;
    gap[0] = 1;
    while( !q.empty() ) {
        int u = q.front();
        q.pop();
        for( int i = head[u]; ~i; i = e[i].nxt ) {
            int v = e[i].v;
            if( ~dep[v] )
                continue;
            dep[v] = dep[u] + 1;
            gap[dep[u]]++;
            q.push( v );
        }
    }
}

int isap()
{
    memcpy( cur, head,  sizeof cur );
    rev_bfs();
    int flow = 0, i, u = st, nv = ed+1;
    top = 0;
    while( dep[st] < nv ) {
        if( u == ed ) {
            int tmp, minn = inf;
            for( i = 0; i < top; ++i ) {
                if( minn > e[s[i]].w ) {
                    minn = e[s[i]].w;
                    tmp = i;
                }
            }
            for( i = 0; i < top; ++i ) {
                e[s[i]].w -= minn;
                e[s[i]^1].w += minn;
            }
            flow += minn;
            top = tmp;
            u = e[s[top]^1].v;
        }
        for( i = cur[u]; ~i; i = e[i].nxt ) {
            if( e[i].w > 0 && dep[u] == dep[e[i].v] + 1 ) {
                cur[u] = i;
                break;
            }
        }
        if( ~i ) {
            s[top++] = cur[u];
            u = e[i].v;
        }
        else {
            if( 0 == (--gap[dep[u]]) )
                break;
            int minn = nv;
            for( i = head[u]; ~i; i = e[i].nxt ) {
                int v = e[i].v;
                if( e[i].w > 0 && minn > dep[v] ) {
                    minn = dep[v];
                    cur[u] = i;
                }
            }
            dep[u] = minn + 1;
            gap[dep[u]]++;
            if( u != st ) {
                u = e[ s[--top]^1 ].v;
            }
        }
    }
    return flow;
}

int main()
{
    int tot;
    scanf("%d", &tot);
    for( int ca = 1; ca <= tot; ca++ ) {
        init();
        scanf("%d%d", &n, &k);
        for( int i = 1; i <= n; ++i )
            scanf("%s", a[i]+1);
        for( int i = 1; i <= n; ++i )
            scanf("%s", b[i]+1);
        m = strlen( a[1]+1 );
        int p = 0, sum = 0;
        for( int i = 1; i <= n; ++i ) {
            for( int j = 1; j <= m; ++j ) {
                if( a[i][j] >= '1' && a[i][j] <= '9' ) {
                    tid[i][j] = ++p;        // re_id
                    add( 2*p-1, 2*p, a[i][j] - '0' );
                }
                if( b[i][j] == 'L' ) {
                    sum++;
                    add( st, 2*tid[i][j]-1, 1 );
                }
            }
        }
        for( int i = 1; i <= n; ++i ) {
            for( int j = 1; j <= m; ++j ) {
                if( !tid[i][j] )
                    continue;
                for( int e = i-k; e <= i+k; ++e ) {
                    for( int r = j-k; r <= j+k; ++r ) {
                        if( !e || !r || e > n || r > m || ( e == i && r == j ) )
                            continue;
                        if( !tid[e][r] || dis( i, j, e, r ) > k )
                            continue;
                        add( 2*tid[i][j], 2*tid[e][r]-1, inf );
                    }
                }
                if( i - k <= 0 || j - k <= 0 || i+k > n || j+k > m )
                    add( 2*tid[i][j], ed, inf );
            }
        }
        int ans = sum - isap();
        if( ans == 0 )
            printf("Case #%d: no lizard was left behind.\n", ca);
        else if( ans == 1)
            printf("Case #%d: 1 lizard was left behind.\n", ca);
        else
            printf("Case #%d: %d lizards were left behind.\n", ca, ans );
        //printf("%d %d\n", sum, ans);
    }
    //system("pause");
    return 0;
}