【网络流】【最小割】洛谷P2598 [ZJOI2009]狼和羊的故事

（毛绒绒的题）忽然就想起beastars。
题目 ------Link------

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题目

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解

最小割=网络流中的最大流。
建图：
S -> 羊：边权 = INF
羊->空地，羊->狼，空地->狼：边权 = 1
狼->T：边权 = INF

然后网络流跑个最大流即可。

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代码

#include
#include
#include
#include
#include
#define innf 1000000000
using namespace std;
int n,m,s,t,tot,lx[101][101],B[100001],dis[100001],l[100001];
int fx[6] = {0,0,+1,-1}, fy[6] = {+1,-1,0,0};
struct asdf{
 int to, next, sy, ff;
} a[10000001];
void add(int xx, int yy, int zz){  //建边
 a[++tot] = (asdf){yy, l[xx], zz, tot + 1}; l[xx] = tot;
 a[++tot] = (asdf){xx, l[yy],  0, tot - 1}; l[yy] = tot;
}
bool spfa(){  //寻找增广路
 queue<int> Q;
 memset(B, 0, sizeof(B));
 memset(dis, 0x7f, sizeof(dis));
 dis[s] = 0; B[s] = 1;
 Q.push(s);
 while(Q.size()){
  int h = Q.front(); 
  Q.pop();
  B[h] = 0;
  for(int i = l[h]; i; i = a[i].next)
    if(a[i].sy && dis[h] + 1 < dis[a[i].to]){
       dis[a[i].to] = dis[h] + 1;
       if(a[i].to == t) return 1;
       if(B[a[i].to] == 0){
          B[a[i].to] = 1;
          Q.push(a[i].to);
       }
    }
 } 
 return 0;
}
int dfs(int d, int minn){  //增广
 if(d == t) return minn;
 int ljl = 0;
 for(int i = l[d]; i; i = a[i].next)
   if(a[i].sy && dis[d] + 1 == dis[a[i].to]){
     int ll = dfs(a[i].to, min(minn - ljl, a[i].sy));
     a[i].sy -= ll;
     a[a[i].ff].sy += ll;
     ljl += ll;
     if(ljl == minn) return ljl;
   }
 return ljl;
}
int dinic(){  //网络流
 int ans = 0;
 while(spfa()) ans += dfs(s, innf);
 return ans;
} 
int main(){
 scanf("%d%d", &n, &m);
 s = 0; t = n * m + 1;
 for(int i = 1; i <= n; ++i)
   for(int j = 1; j <= m; ++j){  //与s，t建边
      scanf("%d", &lx[i][j]);
      if(lx[i][j] == 1) add(s, (i - 1) * m + j, innf);
      if(lx[i][j] == 2) add((i - 1) * m + j, t, innf);
   }     
 for(int i = 1; i <= n; ++i)
   for(int j = 1; j <= m; ++j)
   if(lx[i][j] < 2){
      int bh1 = (i - 1) * m + j;
      for(int k = 0; k < 4; ++k){  //向四周连边
         int xx = i + fx[k], yy = j + fy[k], bh2 = (xx - 1) * m + yy;
         if(xx < 1 || xx > n || yy < 1 || yy > m) continue;
         if(lx[i][j] == 0 && lx[xx][yy] != 1) add(bh1, bh2, 1);
         if(lx[i][j] == 1 && lx[xx][yy] != 1) add(bh1, bh2, 1);
      }
   }
 printf("%d", dinic());
}