HDU3046 Pleasant sheep and big big wolf

Pleasant sheep and big big wolf
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2928 Accepted Submission(s): 1199

Problem Description
In ZJNU, there is a well-known prairie. And it attracts pleasant sheep and his companions to have a holiday. Big big wolf and his families know about this, and quietly hid in the big lawn. As ZJNU ACM/ICPC team, we have an obligation to protect pleasant sheep and his companions to free from being disturbed by big big wolf. We decided to build a number of unit fence whose length is 1. Any wolf and sheep can not cross the fence. Of course, one grid can only contain an animal.
Now, we ask to place the minimum fences to let pleasant sheep and his Companions to free from being disturbed by big big wolf and his companions.

Input
There are many cases.
For every case:

N and M（N,M<=200）
then N*M matrix:
0 is empty, and 1 is pleasant sheep and his companions, 2 is big big wolf and his companions.

Output
For every case:

First line output “Case p:”, p is the p-th case;
The second line is the answer.

Sample Input
4 6
1 0 0 1 0 0
0 1 1 0 0 0
2 0 0 0 0 0
0 2 0 1 1 0

Sample Output
Case 1:
4

把羊和狼分割，计算最小割
1是羊，2是狼，0是墙
用dinic算法，先bfs建立层次图，然后dfs找出增广路，直到没有增广为止，建图用邻接表

#include <cstring>#include <cstdlib>#include <cstring>#include <algorithm>#include <cstdio>#include <queue>#define RE(x) (x)^1#define INF 0x3fffffff#define MAXN 205using namespace std;int N, M, G[MAXN][MAXN], dis[MAXN*MAXN];int dir[2][2] = {0, 1, 1, 0};int idx, head[MAXN*MAXN];const int sink = MAXN*MAXN-1, source = MAXN*MAXN-2;struct Edge{    int v, cap, next;} e[MAXN*MAXN*4];void init(){    idx = 0;    memset(head, 0xff, sizeof (head));}int to(int x, int y){    return (x-1)*M+(y-1);}void insert(int u,int v,int w){    e[idx].v=v,e[idx].cap=w;    e[idx].next=head[u];    head[u]=idx;    idx++;}bool judge(int x, int y)//判断越界{    if (x < 1 || x > N || y < 1 || y > M)    {        return false;    }    return true;}void check(int x, int y){    int xx, yy;    if (G[x][y] == 1)    {        insert(to(x, y), sink, INF);        insert(sink, to(x, y), 0);    }    else if (G[x][y] == 2)    {        insert(source, to(x, y), INF);        insert(to(x, y), source, 0);    }    for (int i = 0; i < 2; ++i)    {        xx = x + dir[i][0], yy = y + dir[i][1];        if (judge(xx, yy))        {            insert(to(x, y), to(xx, yy), 1);            insert(to(xx, yy), to(x, y), 1);        }    }}bool bfs()//寻找深度图，dis代表层次{    int pos;    queue<int>q;    memset(dis, 0xff, sizeof (dis));    dis[source] = 0;    q.push(source);    while (!q.empty())    {        pos = q.front();        q.pop();        for (int i = head[pos]; i != -1; i = e[i].next)        {            if (dis[e[i].v] == -1 && e[i].cap > 0)            {                dis[e[i].v] = dis[pos]+1;                q.push(e[i].v);            }        }    }    //return dis[sink] != -1;    if(dis[sink]!=-1)    return 1;    else    return 0;}int dfs(int u,int flow){    if(u==sink)    return flow;    int sf=0,fs;    for(int i=head[u];i!=-1;i=e[i].next)    {        if(dis[u]+1==dis[e[i].v]&&e[i].cap>0&&(fs=dfs(e[i].v,min(flow-sf,e[i].cap))))        {            e[i].cap-=fs;            e[i^1].cap+=fs;            sf+=fs;            if(sf==flow)            return flow;        }    }    if(!sf)    dis[u]=-1;    return sf;}int dinic(){    int ans = 0;    while (bfs())    {        ans += dfs(source, INF);    }    return ans;}int main(){    int ca = 0;    while (scanf("%d %d", &N, &M) == 2)    {        init();        for (int i = 1; i <= N; ++i)        {            for (int j = 1; j <= M; ++j)            {                scanf("%d", &G[i][j]);                //    getint(G[i][j]);            }        }        for (int i = 1; i <= N; ++i)        {            for (int j = 1; j <= M; ++j)            {                check(i, j);            }        }        printf("Case %d:\n%d\n", ++ca, dinic());    }    return 0;}