HDU 4859 (bestcoder #1 1003) coastline (minimum cut of network flow)

Address: hdu4859

After doing multiple schools in Hangzhou and Hangzhou, there will be too few knowledge points. We should focus on the topic-based supplementary knowledge points. Next year's multiple schools will be the focus.

The longest perimeter of this question, you can imagine that the coastline here must be ". "and" D ", that is, find the most adjacent". and D pairs. You can first convert it to the smallest cut to find the smallest pair logarithm. Because the total logarithm is certain, you only need to subtract it.

First, we need to fill the "D" in the previous circle, and then it becomes a rectangle (n + 2) * (m + 2. Because the numbers of adjacent matches are required, you can use black and white dyeing to convert them into a bipartite graph model and add edges to each adjacent graph. The bipartite graph has been divided into two child sets: X and Y. Assume that all the ". "Both are allocated to X sets, and all" D "must be allocated to y sets (only assumptions). At this time, we will discuss this in two cases:

1: if the current point is on the map but is divided into the y set, or the current point is D, but is divided into the X set, a INF edge is connected to the source point.

2: If the current point is on the map and is divided into X sets, or the current point is d, and is divided into y sets, a inf edge is connected to the sink point.

After analysis, we can see that only. -->. Or d --> D, which is of the same type, can be remitted from the source stream.

The source connection point is the "Incorrect" Point (set D or set Y of X .), the point of the sink connection is a "correct" Point (of the X set. or D of the Y set. Therefore, if you want to migrate data from the source stream to the sink, it must be d-> D or. -> ., that is, for the same type, the minimum cut is to minimize the same.

And then subtract the total logarithm. This is the answer.

The Code is as follows:

#include <iostream>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;int head[3000], source, sink, nv, cnt;int cur[3000], num[3000], d[3000], pre[3000];int jx[]={0,0,1,-1};int jy[]={1,-1,0,0};struct node{    int u, v, cap, next;}edge[1000000];void add(int u, int v, int cap){    edge[cnt].v=v;    edge[cnt].cap=cap;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].cap=0;    edge[cnt].next=head[v];    head[v]=cnt++;}void bfs(){    memset(d,-1,sizeof(d));    memset(num,0,sizeof(num));    queue<int>q;    q.push(sink);    d[sink]=0;    num[0]=1;    while(!q.empty())    {        int u=q.front();        q.pop();        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v;            if(d[v]==-1)            {                d[v]=d[u]+1;                num[d[v]]++;                q.push(v);            }        }    }}int isap(){    memcpy(cur,head,sizeof(cur));    bfs();    int flow=0, u=pre[source]=source, i;    while(d[source]<nv)    {        if(u==sink)        {            int f=INF, pos;            for(i=source;i!=sink;i=edge[cur[i]].v)            {                if(f>edge[cur[i]].cap)                {                    f=edge[cur[i]].cap;                    pos=i;                }            }            for(i=source;i!=sink;i=edge[cur[i]].v)            {                edge[cur[i]].cap-=f;                edge[cur[i]^1].cap+=f;            }            flow+=f;            u=pos;        }        for(i=cur[u];i!=-1;i=edge[i].next)        {            if(d[edge[i].v]+1==d[u]&&edge[i].cap) break;        }        if(i!=-1)        {            cur[u]=i;            pre[edge[i].v]=u;            u=edge[i].v;        }        else        {            if(--num[d[u]]==0) break;            int mind=nv;            for(i=head[u];i!=-1;i=edge[i].next)            {                if(mind>d[edge[i].v]&&edge[i].cap)                {                    mind=d[edge[i].v];                    cur[u]=i;                }            }            d[u]=mind+1;            num[d[u]]++;            u=pre[u];        }    }    return flow;}int main(){    int t, n, m, i, j, mp[60][60], num=0;    char s[60];    scanf("%d",&t);    while(t--)    {        num++;        scanf("%d%d",&n,&m);        memset(mp,0,sizeof(mp));        for(i=1;i<=n;i++)        {            scanf("%s",s);            for(j=0;j<m;j++)            {                if(s[j]=='E')                    mp[i][j+1]=2;                else if(s[j]=='.')                    mp[i][j+1]=1;            }        }        cnt=0;        memset(head,-1,sizeof(head));        source=0;        sink=(n+2)*(m+2)+1;        nv=sink+1;        for(i=0;i<=n+1;i++)        {            for(j=0;j<=m+1;j++)            {                if((i+j)%2==0)                {                    if(mp[i][j]==1)                    {                        add(i*(m+2)+j+1,sink,INF);                    }                    else if(mp[i][j]==0)                    {                        add(source,i*(m+2)+j+1,INF);                    }                }                else                {                    if(mp[i][j]==0)                    {                        add(i*(m+2)+j+1,sink,INF);                    }                    else if(mp[i][j]==1)                    {                        add(source,i*(m+2)+j+1,INF);                    }                }                for(int k=0;k<4;k++)                {                    int x=i+jx[k];                    int y=j+jy[k];                    if(x>=0&&x<=n+1&&y>=0&&y<=m+1)                    {                        add(i*(m+2)+j+1,x*(m+2)+y+1,1);                    }                }            }        }        int ans;        ans=isap();        printf("Case %d: %d\n",num,(n+1)*(m+2)+(n+2)*(m+1)-ans);    }    return 0;}

HDU 4859 (bestcoder #1 1003) coastline (minimum cut of network flow)