P3308 [SDOI2014]LIS

【题意】

 【分析】

LIS部分和网络流24题中的P2766 最长不下降子序列问题一致

主要是多了一个输出方案，这个就需要用到贪心退流的方式了

 【代码】

#include<bits/stdc++.h>
using namespace std;
#define INF 1e9
const int M=1000000;
const int N=1500;
int tot,nxt[M],point[N],remind[M],v[M],dis[N],cur[N],n,a[N],b[N],f[N],ans[N];
struct hh{int c,bh,x;}li[N];
int read()
{
    int x=0;char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch<='9' && ch>='0') x=x*10+ch-'0',ch=getchar();
    return x;
}
void addline(int x,int y,int z)
{
    ++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remind[tot]=z;
    ++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remind[tot]=0;
}
bool bfs(int s,int t)
{
    for (int i=0;i<=n*2+1;i++) cur[i]=point[i];
    queue<int>q;q.push(s);
    memset(dis,0x7f,sizeof(dis));dis[s]=0;
    while (!q.empty())
    {
        int now=q.front();q.pop();
        for (int i=point[now];i!=-1;i=nxt[i])
          if (dis[v[i]]>INF && remind[i])
          {
            dis[v[i]]=dis[now]+1; q.push(v[i]);
            if (dis[t]<INF) return 1;
          } 
    }
    return 0;
}
int dfs(int now,int t,int limit)
{
    if (now==t || !limit) return limit;
    int flow=0,f;
    for (int i=cur[now];i!=-1;i=nxt[i])
    {
        cur[now]=i;
        if (remind[i] && dis[v[i]]==dis[now]+1)
        {
            f=dfs(v[i],t,min(limit,remind[i]));
            limit-=f; flow+=f;
            remind[i]-=f; remind[i^1]+=f;
            if (!limit) return flow;
        }
    }
    return flow;
}
int dinic(int s,int t)
{
    int ans=0;
    while (bfs(s,t)) ans+=dfs(s,t,INF);
    return ans;
}
int cmp(hh a,hh b){return a.c<b.c;}
int main()
{
    // freopen("a.in","r",stdin);
    // freopen("a.out","w",stdout);
    int T;T=read();
    while (T--)
    {
        tot=-1;memset(point,-1,sizeof(point));
        n=read();int len=0;
        for (int i=1;i<=n;i++) a[i]=read();
        for (int i=1;i<=n;i++) b[i]=read();
        for (int i=1;i<=n;i++) li[i].c=read(),li[i].x=i;
        for (int i=1;i<=n;i++)
        {
            int maxx=0;
            for (int j=1;j<i;j++)
              if (a[j]<a[i] && f[maxx]<f[j]) maxx=j;
            f[i]=f[maxx]+1;
            len=max(len,f[i]);
        }
        int s=0,t=n*2+1;
        for (int i=1;i<=n;i++)
        {
            li[i].bh=tot+1,addline(i,i+n,b[i]);
            if (f[i]==1) addline(s,i,INF);
            else if (f[i]==len) addline(i+n,t,INF);
            for (int j=i+1;j<=n;j++) 
              if (a[j]>a[i] && f[j]==f[i]+1) addline(i+n,j,INF);
        }
        int m=dinic(s,t),num=0;
        printf("%d ",m);
        sort(li+1,li+n+1,cmp);
        for (int i=1;i<=n;i++)
        {
            if (bfs(li[i].x,li[i].x+n)) continue; 
            dinic(t,li[i].x+n); dinic(li[i].x,s);
            ans[++num]=li[i].x;
            remind[li[i].bh]=remind[li[i].bh^1]=0;
        }
        sort(ans+1,ans+num+1);printf("%d\n%d",num,ans[1]);
        for (int i=2;i<=num;i++) printf(" %d",ans[i]);
        printf("\n");
    }
}

标签：nxt,ch,P3308,int,point,tot,SDOI2014,LIS,dis
来源： https://www.cnblogs.com/andylnx/p/14850841.html