HDU 3046 Pleasant sheep and big big wolf

Pleasant sheep and big big wolf

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3046
64-bit integer IO format: %I64d      Java class name: Main

 

 

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

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

 

解题：最小割。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 210*210;
struct arc{
    int to,flow,next;
    arc(int x = 0,int y = 0,int z = -1){
        to = x;
        flow = y;
        next = z;
    }
};
arc e[maxn<<2];
int head[maxn],d[maxn],cur[maxn];
int tot,S,T,n,m;
void add(int u,int v,int flow){
    e[tot] = arc(v,flow,head[u]);
    head[u] = tot++;
    e[tot] = arc(u,flow,head[v]);//可以不这么建图。。。。
    head[v] = tot++;
}
bool bfs(){
    memset(d,-1,sizeof(d));
    queue<int>q;
    q.push(T);
    d[T] = 1;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = e[i].next){
            if(e[i^1].flow && d[e[i].to] == -1){
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
    }
    return d[S] > -1;
}
int dfs(int u,int low){
    if(u == T) return low;
    int tmp = 0,a;
    for(int &i = cur[u]; ~i; i = e[i].next){
        if(e[i].flow && d[e[i].to]+1==d[u]&&(a=dfs(e[i].to,min(low,e[i].flow)))){
            e[i].flow -= a;
            e[i^1].flow += a;
            low -= a;
            tmp += a;
            if(!low) break;
        }
    }
    if(!tmp) d[u] = -1;
    return tmp;
}
int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(cur,head,sizeof(head));
        ans += dfs(S,INF);
    }
    return ans;
}
int main() {
    int u,v,w,cs = 1;
    while(~scanf("%d %d",&n,&m)){
        memset(head,-1,sizeof(head));
        S = n*m;
        T = n*m+1;
        for(int i = tot = 0; i < n; ++i)
        for(int j = 0; j < m; ++j){
            scanf("%d",&u);
            if(i) add(m*(i-1)+j,m*i+j,1);
            if(j) add(m*i+j-1,m*i+j,1);
            if(u == 1) add(S,m*i+j,INF);
            if(u == 2) add(m*i+j,T,INF);
        }
        printf("Case %d:
%d
",cs++,dinic());
    }
    return 0;
}