[CF Gym 100827C] Containment [2014-2015 ACM-ICPC Pacific Northwest Regional Contest C]

题意

三维空间10*10*10方块，一些方块坏掉了，求用最少的面包裹住坏掉的方块，可以同时包裹好的方块。其中面只能处于方块间隙中。

题解

典型的最小割问题。以每个方块为点，S与最外围所有点连边，其中在顶点处的点连3边，不在顶点且在棱上的点连2边，其他点连1边。好点与周围六个方向点连1边，坏点与T连inf边，求最小割即可。

代码

/****************************************\* Author : ztx* Title  : C - Containment* ALG    : 网络流-最小割* CMT    : 能割的地方连流量为1的边，S与最外围所有点连1边，黑色点之间连inf边，黑色点与T连inf边，求最小割* Time   :\****************************************/#include <cstdio>#define Rep(i,l,r) for(i=(l);i<=(r);i++)#define rep(i,l,r) for(i=(l);i< (r);i++)#define Rev(i,r,l) for(i=(r);i>=(l);i--)#define rev(i,r,l) for(i=(r);i> (l);i--)typedef long long ll ;typedef double lf ;int CH , NEG ;template <typename TP>inline void read(TP& ret) {    ret = NEG = 0 ; while (CH=getchar() , CH<'!') ;    if (CH == '-') NEG = true , CH = getchar() ;    while (ret = ret*10+CH-'0' , CH=getchar() , CH>'!') ;    if (NEG) ret = -ret ;}template <typename TP>inline void readc(TP& ret) {    while (ret=getchar() , ret<'!') ;    while (CH=getchar() , CH>'!') ;}template <typename TP>inline void reads(TP *ret) {    ret[0]=0;while (CH=getchar() , CH<'!') ;    while (ret[++ret[0]]=CH,CH=getchar(),CH>'!') ;    ret[ret[0]+1]=0;}#include <cstring>#define  maxn  15LL#define  maxN  10000LL#define  maxM  100000LL#define  infi  0x3f3f3f3fLLstruct FST { int to , next , flow ; } e[maxM<<1] ;int star[maxN] , tote ;inline void FST_init() { memset(star , 0 , sizeof star) ; tote = 1 ; }inline void AddEdge(int u , int v , int cap) {    e[++tote].to = v ; e[tote].flow = cap ; e[tote].next = star[u] ; star[u] = tote ;    e[++tote].to = u ; e[tote].flow = 0 ; e[tote].next = star[v] ; star[v] = tote ;}#define  min(x,y) ((x)<(y)?(x):(y))int N , S , T ;int h[maxN] , vh[maxN] ;int dfs(int u , int flowu) {int p , tmp = h[u]+1 ;int flow = 0 , flowv ;    if (u == T) return flowu ;    for (p = star[u] ; p ; p = e[p].next) {        if (e[p].flow && (h[e[p].to]+1==h[u])) {            flowv = dfs(e[p].to , min(flowu-flow , e[p].flow)) ;            flow += flowv ; e[p].flow -= flowv ; e[p^1].flow += flowv ;            if (flow==flowu || h[S]==N) return flow ;        }    }    for (p = star[u] ; p ; p = e[p].next)        if (e[p].flow) tmp = min(tmp , h[e[p].to]) ;    if (--vh[h[u]] == 0) h[S] = N ;    else ++ vh[h[u]=tmp+1] ;    return flow ;}int SAP() {    int ret = 0 ;    memset(vh , 0 , sizeof vh) ;    memset(h , 0 , sizeof h) ;    vh[S] = N ;    while (h[S] < N) ret += dfs(S , infi) ;    return ret ;}bool g[maxn][maxn][maxn] ;int f[maxn][maxn][maxn] ;const int dx[6] = {0,0,1,-1,0,0} ;const int dy[6] = {0,0,0,0,1,-1} ;const int dz[6] = {1,-1,0,0,0,0} ;int main() {int Time , i , j , k , ii , jj , kk , dir , m ;//    #define READ    #ifdef  READ        freopen(".in" ,"r",stdin ) ;        freopen(".out","w",stdout) ;    #endif    N = 0 ;    Rep (i,1,10) Rep (j,1,10) Rep (k,1,10)        f[i][j][k] = ++N ;    S = ++N , T = ++N ;    for (read(Time) ; Time -- ; ) {        read(m) ;        memset(g,0,sizeof g) ;        FST_init() ;        Rep (i,1,m) {            read(ii) , read(jj) , read(kk) ; // 0->9  ==>  1->10            g[ii+1][jj+1][kk+1] = 1 ;        }        Rep (i,1,10) Rep (j,1,10) Rep (k,1,10) {            dir = 0 ;            if (i==1||i==10) dir ++ ;            if (j==1||j==10) dir ++ ;            if (k==1||k==10) dir ++ ;            if (dir) AddEdge(S,f[i][j][k],dir) ;            if (g[i][j][k]) AddEdge(f[i][j][k],T,infi) ;            else {                rep (dir,0,6) {                    ii = i+dx[dir] ;                    jj = j+dy[dir] ;                    kk = k+dz[dir] ;                    if (ii>0&&ii<11&&jj>0&&jj<11&&kk>0&&kk<11) {                        AddEdge(f[i][j][k],f[ii][jj][kk],1) ;                    }                }            }        }        printf("%d\n", SAP()) ;    }    #ifdef  READ        fclose(stdin) ; fclose(stdout) ;    #else        getchar() ; getchar() ;    #endif    return 0 ;}