【BZOJ 1412】[ZJOI2009]狼和羊的故事

【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

每个点和相邻的4个点连一条容量为1的边。 然后源点和每头羊连容量INF的边 每头狼和汇点连容量INF的边。 这样求最小割的时候只会把栅栏删掉。 然后源点不能到汇点了。 显然就是每头羊都不能和狼联通了(否则肯定能有增广路

a数组一开始写成1万x1万。
(直接超时了。。

【代码】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define all(x) x.begin(),x.end()
#define pb push_back
#define ms(x,y) memset(x,y,sizeof x)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
 
const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0};
const int N = 10000;
const int INF = 10000+10;
 
struct abc{
    int en,nex;
    LL flow;
};
 
int n,m,s,t,cost[N+10],deep[N*2+20];
int fir[N+10],tfir[N+10],totm;
abc bian[2*N+2*N+N*8+10];
queue <int> dl;
 
void add(int x,int y,LL cost){
    bian[totm].nex = fir[x];
    fir[x] = totm;
    bian[totm].en = y,bian[totm].flow = cost;
    totm++;
 
    bian[totm].nex = fir[y];
    fir[y] = totm;
    bian[totm].en = x,bian[totm].flow = 0;
    totm++;
}
 
bool bfs(int s,int t){
    dl.push(s);
    ms(deep,255);
    deep[s] = 0;
 
    while (!dl.empty()){
        int x = dl.front();
        dl.pop();
        for (int temp = fir[x]; temp!= -1 ;temp = bian[temp].nex){
            int y = bian[temp].en;
            if (deep[y]==-1 && bian[temp].flow>0){
                deep[y] = deep[x] + 1;
                dl.push(y);
            }
        }
    }
    return deep[t]!=-1;
}
 
LL dfs(int x,int t,LL limit){
    if (x == t) return limit;
    if (limit == 0) return 0;
    LL cur,f = 0;
    for (int temp = tfir[x];temp!=-1;temp = bian[temp].nex){
        tfir[x] = temp;
        int y = bian[temp].en;
        if (deep[y] == deep[x] + 1 && (cur = dfs(y,t,min(limit,(LL)bian[temp].flow))) ){
            f += cur;
            limit -= cur;
            bian[temp].flow -= cur;
            bian[temp^1].flow += cur;
            if (!limit) break;
        }
    }
    return f;
}
int a[100+10][100+10];
 
int main(){
    #ifdef LOCAL_DEFINE
        freopen("rush_in.txt", "r", stdin);
    #endif
    ms(fir,255);
    scanf("%d%d",&n,&m);
    for (int i = 1;i <= n;i++)
        for (int j = 1;j <= m;j++)
            scanf("%d",&a[i][j]);
 
    for (int i = 1;i <= n;i++){
        for (int j = 1;j <= m;j++){
            for (int k = 0;k < 4;k++){
                int ti = i+dx[k],tj = j+dy[k];
                if (ti>=1 && ti<=n && tj>=1 && tj<=m){
                    add((i-1)*m+j,(ti-1)*m+tj,1);
                }
            }
        }
    }
 
    for (int i = 1;i <= n;i++)
        for (int j = 1;j <=m;j++){
            if (a[i][j]==1){
                add(0,(i-1)*m+j,INF);
            }else if (a[i][j]==2){
                add((i-1)*m+j,n*m+1,INF);
            }
        }
 
    s = 0,t = n*m+1;
    int ans = 0;
    while ( bfs(s,t) ){
        for (int i = 0;i <= n*m+1;i++) tfir[i] = fir[i];
        ans += dfs(s,t,INF);
    }
    printf("%d\n",ans);
    return 0;
}
posted @ 2018-05-25 15:25  AWCXV  阅读(102)  评论(0编辑  收藏  举报