A. Cutting Figure
You've gotten an n × m sheet of squared paper. Some of its squares are painted. Let's mark the set of all painted squares as A. Set A is connected. Your task is to find the minimum number of squares that we can delete from set A to make it not connected.
A set of painted squares is called connected, if for every two squares a and b from this set there is a sequence of squares from the set, beginning in a and ending in b, such that in this sequence any square, except for the last one, shares a common side with the square that follows next in the sequence. An empty set and a set consisting of exactly one square are connected by definition.
The first input line contains two space-separated integers n and m (1 ≤ n, m ≤ 50) — the sizes of the sheet of paper.
Each of the next n lines contains m characters — the description of the sheet of paper: the j-th character of the i-th line equals either "#", if the corresponding square is painted (belongs to set A), or equals "." if the corresponding square is not painted (does not belong to set A). It is guaranteed that the set of all painted squares A is connected and isn't empty.
On the first line print the minimum number of squares that need to be deleted to make set A not connected. If it is impossible, print -1.
5 4
####
#..#
#..#
#..#
####
2
5 5
#####
#...#
#####
#...#
#####
2
In the first sample you can delete any two squares that do not share a side. After that the set of painted squares is not connected anymore.
The note to the second sample is shown on the figure below. To the left there is a picture of the initial set of squares. To the right there is a set with deleted squares. The deleted squares are marked with crosses.
View Code
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; #define V 2550 #define E 300000 #define inf 0xffff int map[V][2]; int flag[V][V]; char g[55][55]; int log[55][55]; struct Edge { int u,v,c,next; }edge[E]; int n,m,cnt; int dist[V]; int head[V]; int que[V]; int sta[V]; int s,t; void init(){ cnt=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int c){ edge[cnt].u=u;edge[cnt].v=v;edge[cnt].c=c; edge[cnt].next=head[u];head[u]=cnt++; edge[cnt].u=v;edge[cnt].v=u;edge[cnt].c=0; edge[cnt].next=head[v];head[v]=cnt++; } int dinic(int s,int t){ int ans=0; while(true){ int left,right,u,v; memset(dist,-1,sizeof(dist)); left=right=0; que[right++]=s; dist[s]=0; while(left<right){ u=que[left++]; for(int k=head[u];k!=-1;k=edge[k].next){ u=edge[k].u; v=edge[k].v; if(edge[k].c > 0 && dist[v]==-1){ dist[v]=dist[u]+1; que[right++]=v; if(v==t){ left=right; break; } } } } if(dist[t]==-1) break; int top=0; int now=s; while(true){ if(now!=t){ int k; for(k=head[now];k!=-1;k=edge[k].next){ if(edge[k].c>0 && dist[edge[k].v]==dist[edge[k].u]+1) break; } if(k!=-1){ sta[top++]=k; now=edge[k].v; } else{ if(top==0) break; dist[edge[sta[--top]].v]=-1; now=edge[sta[top]].u; } } else{ int flow=inf,ebreak; for(int i=0;i<top;i++){ if(flow>edge[sta[i]].c){ flow=edge[sta[i]].c; ebreak=i; } } ans+=flow; for(int i=0;i<top;i++){ edge[sta[i]].c-=flow; edge[sta[i]^1].c+=flow; } now=edge[sta[ebreak]].u; top=ebreak; } } } return ans; } void build(int x,int y,int ver,int n,int m){ init(); for(int i=1;i<=n;i++){ addedge(i,i+n,1); } for(int i=0;i<m;i++){ addedge(map[i][0]+n,map[i][1],inf); addedge(map[i][1]+n,map[i][0],inf); } addedge(x,x+n,inf); addedge(y,y+n,inf); } int main() { //freopen("in.txt","r",stdin); while(scanf("%d%d",&n,&m)!=EOF){ for(int i=0;i<n;i++){ scanf("%s",g[i]); } memset(log,0,sizeof(log)); int cnt=1; for(int i=0;i<n;i++) for(int j=0;j<m;j++) if(g[i][j]=='#'){ log[i][j]=cnt++; } else log[i][j]=0; /*for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ cout <<log[i][j]; } cout <<endl; }*/ int ver=(cnt-1)*2+1; //cout << ver << endl; int k=0,a,b; memset(map,0,sizeof(map)); memset(flag,0,sizeof(flag)); for(int i=0;i<n;i++) for(int j=0;j<m;j++) if(log[i][j]){ if(log[i+1][j]){ a=map[k][0]=log[i][j]; b=map[k][1]=log[i+1][j]; flag[a][b]=1; //cout << a <<"--" << b << endl; k++; } if(log[i-1][j]){ a=map[k][0]=log[i][j]; b=map[k][1]=log[i-1][j]; flag[a][b]=1; //cout << a <<"--" << b << endl; k++; } if(log[i][j-1]){ a=map[k][0]=log[i][j]; b=map[k][1]=log[i][j-1]; flag[a][b]=1; //cout << a <<"--" << b << endl; k++; } if(log[i][j+1]){ a=map[k][0]=log[i][j]; b=map[k][1]=log[i][j+1]; flag[a][b]=1; //cout << a <<"--" << b << endl; k++; } } /*for(int i=0;i<m;i++){ scanf(" (%d,%d)",&a,&b); a++,b++; map[i][0]=a,map[i][1]=b; flag[a][b]=1; }*/ if(k==0){ if((cnt-1)==1) printf("-1\n"); else printf("0\n"); continue; } //cout << k << endl; int pre,ans=inf,sign=0; for(int i=1;i<=cnt-1;i++){ for(int j=i+1;j<=cnt-1;j++){ if(!flag[i][j]){ sign=1; build(i,j,ver,cnt-1,k); ans=min(ans,dinic(i,j+cnt-1)); } } } if(sign){ printf("%d\n",ans); } else{ printf("-1\n"); } } return 0; }
正解应该用dfs爆之,涂方便,直接用“网络流求割顶集模板n^4”开敲,果断TLE。。。尽情来鄙视我吧。。。
