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  • BZOJ 1934 善意的投票(最小割)

    把人分成两个集合,一个赞成睡觉,一个反对睡觉。好朋友连一条容量为1的双向边,s向赞成睡觉的连边,反对睡觉的向t连边。

    那么这个图的一个割就对应着一个方案。如果割掉s和v的边,就代表v投意见与它自己相反的票,t和v的边同理。割掉u和v的边,就代表了这对好朋友之间意见不同。

    这样求出一个割之后,好朋友之间意见不同的边都被割去了。

    求出此图的最小割即为答案。

    # include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi acos(-1.0)
    # define eps 1e-9
    # define MOD 1024523
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=305;
    //Code begin...
    
    struct Edge{int p, next, w;}edge[200005];
    int head[N], cnt=2, s, t, vis[N];
    queue<int>Q;
    
    void add_edge(int u, int v, int w){
        edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;
        edge[cnt].p=u; edge[cnt].w=0; edge[cnt].next=head[v]; head[v]=cnt++;
    }
    int bfs(){
        int i, v;
        mem(vis,-1);
        vis[s]=0; Q.push(s);
        while (!Q.empty()) {
            v=Q.front(); Q.pop();
            for (i=head[v]; i; i=edge[i].next) {
                if (edge[i].w>0 && vis[edge[i].p]==-1) {
                    vis[edge[i].p]=vis[v] + 1;
                    Q.push(edge[i].p);
                }
            }
        }
        return vis[t]!=-1;
    }
    int dfs(int x, int low){
        int i, a, temp=low;
        if (x==t) return low;
        for (i=head[x]; i; i=edge[i].next) {
            if (edge[i].w>0 && vis[edge[i].p]==vis[x]+1){
                a=dfs(edge[i].p,min(edge[i].w,temp));
                temp-=a; edge[i].w-=a; edge[i^1].w += a;
                if (temp==0) break;
            }
        }
        if (temp==low) vis[x]=-1;
        return low-temp;
    }
    int main ()
    {
        int n, m, flag, u, v;
        scanf("%d%d",&n,&m); s=0; t=n+1;
        FOR(i,1,n) {
            scanf("%d",&flag);
            if (flag) add_edge(s,i,1);
            else add_edge(i,t,1);
        }
        FOR(i,1,m) scanf("%d%d",&u,&v), add_edge(u,v,1), add_edge(v,u,1);
        int sum=0, tmp;
        while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp;
        printf("%d
    ",sum);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6797225.html
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