[BZOJ3158]千钧一发（最小割）

题目描述

传送门

题解

每个点拆点xi,yi,s->xi,yi->t,bi
对于不能同时选的两个点ij，xi->yj,inf，xj->yi,inf
答案为 (2∗∑i=1nbi−maxflow)/2
刚开始的时候只对于j>i的连边，然后 ∑i=1nbi−maxflow ，但是wa了，不是很理解为什么…感觉这两种做法是等价的啊…
可能是因为图不是对称的所以会出现割一边的情况？

代码

#include
#include
#include
#include
#include
#include
using namespace std;
#define LL long long
#define N 2005
#define inf 2000000000

int n,b,s,t,sum,maxflow;
int tot,point[N],nxt[N*N],v[N*N],remain[N*N];
int deep[N],last[N],num[N],cur[N];
LL a[N];
queue <int> q;

void addedge(int x,int y,int cap)
{
    ++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;
    ++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;
}
void bfs(int t)
{
    for (int i=s;i<=t;++i) deep[i]=t;
    deep[t]=0;
    for (int i=s;i<=t;++i) cur[i]=point[i];
    while (!q.empty()) q.pop();
    q.push(t);

    while (!q.empty())
    {
        int now=q.front();q.pop();
        for (int i=point[now];i!=-1;i=nxt[i])
            if (deep[v[i]]==t&&remain[i^1])
            {
                deep[v[i]]=deep[now]+1;
                q.push(v[i]);
            }
    }
}
int addflow(int s,int t)
{
    int now=t,ans=inf;
    while (now!=s)
    {
        ans=min(ans,remain[last[now]]);
        now=v[last[now]^1];
    }
    now=t;
    while (now!=s)
    {
        remain[last[now]]-=ans;
        remain[last[now]^1]+=ans;
        now=v[last[now]^1];
    }
    return ans;
}
void isap(int s,int t)
{
    bfs(t);
    for (int i=s;i<=t;++i) ++num[deep[i]];

    int now=s;
    while (deep[s]if (now==t)
        {
            maxflow+=addflow(s,t);
            now=s;
        }

        bool has_find=false;
        for (int i=cur[now];i!=-1;i=nxt[i])
            if (deep[v[i]]+1==deep[now]&&remain[i])
            {
                has_find=true;
                cur[now]=i;
                last[v[i]]=i;
                now=v[i];
                break;
            }

        if (!has_find)
        {
            int minn=t-1;
            for (int i=point[now];i!=-1;i=nxt[i])
                if (remain[i]) minn=min(minn,deep[v[i]]);
            if (!(--num[deep[now]])) break;
            ++num[deep[now]=minn+1];
            cur[now]=point[now];
            if (now!=s) now=v[last[now]^1];
        }
    }
}
bool check1(LL a,LL b)
{
    LL T=a*a+b*b;
    LL sT=sqrt(T);
    if (sT*sT==T) return 0;
    else return 1;
}
LL gcd(LL a,LL b)
{
    if (!b) return a;
    else return gcd(b,a%b);
}
bool check2(LL a,LL b)
{
    LL T=gcd(a,b);
    if (T>1) return 1;
    else return 0;
}
int main()
{
    tot=-1;memset(point,-1,sizeof(point));
    scanf("%d",&n);
    s=1,t=n+n+2;
    for (int i=1;i<=n;++i) scanf("%lld",&a[i]);
    for (int i=1;i<=n;++i)
    {
        scanf("%d",&b);
        addedge(s,1+i,b);
        addedge(1+n+i,t,b);
        sum+=b*2;
    }
    for (int i=1;i<=n;++i)
        for (int j=1;j<=n;++j)
        {
            bool flag=check1(a[i],a[j])||check2(a[i],a[j]);
            if (!flag) addedge(1+i,1+n+j,inf);
        }
    isap(s,t);
    printf("%d\n",(sum-maxflow)/2);
}