题目大意:一个农场中有狼和羊,现在要将他们用围栏分开,问最少需要多少围栏。
思路:所有源向所有狼连边,所有羊向汇连边,图中的每个相邻的格子之间连边,然后跑S->T的最大流,也就是把狼和羊分开的最小割。
CODE:
#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 11000 #define MAXE 500010 #define INF 0x3f3f3f3f #define S 0 #define T (m * n + 1) using namespace std; const int dx[] = {0,1,-1,0,0}; const int dy[] = {0,0,0,1,-1}; int m,n; int src[110][110],num[110][110],cnt; int head[MAXE],total = 1; int next[MAXE],aim[MAXE],flow[MAXE]; int deep[MAXE]; inline void Add(int x,int y,int f) { next[++total] = head[x]; aim[total] = y; flow[total] = f; head[x] = total; } inline void Insert(int x,int y,int f) { Add(x,y,f); Add(y,x,0); } inline bool BFS() { static queue<int> q; while(!q.empty()) q.pop(); memset(deep,0,sizeof(deep)); deep[S] = 1; q.push(S); while(!q.empty()) { int x = q.front(); q.pop(); for(int i = head[x]; i; i = next[i]) if(!deep[aim[i]] && flow[i]) { deep[aim[i]] = deep[x] + 1; q.push(aim[i]); if(aim[i] == T) return true; } } return false; } int Dinic(int x,int f) { if(x == T) return f; int temp = f; for(int i = head[x]; i; i = next[i]) if(deep[aim[i]] == deep[x] + 1 && flow[i] && temp) { int away = Dinic(aim[i],min(flow[i],temp)); if(!away) deep[aim[i]] = 0; flow[i] -= away; flow[i^1] += away; temp -= away; } return f - temp; } int main() { cin >> m >> n; for(int i = 1; i <= m; ++i) for(int j = 1; j <= n; ++j) { scanf("%d",&src[i][j]),num[i][j] = ++cnt; if(src[i][j] == 1) Insert(S,num[i][j],INF); if(src[i][j] == 2) Insert(num[i][j],T,INF); } for(int i = 1; i <= m; ++i) for(int j = 1; j <= n; ++j) for(int k = 1; k <= 4; ++k) { int fx = i + dx[k]; int fy = j + dy[k]; if(!num[fx][fy]) continue; Insert(num[i][j],num[fx][fy],1); } int max_flow = 0; while(BFS()) max_flow += Dinic(S,INF); cout << max_flow << endl; return 0; }
BZOJ 1412 ZJOI 2009 狼和羊的故事 最小割
原文地址:http://blog.csdn.net/jiangyuze831/article/details/40820051