【HDOJ6611】K Subsequence（费用流）

题意：给定一个长为n的正整数序列，要求从中取出至多k个不下降序列，使得它们的和最大，求这个和

n<=2e3，k<=10，a[i]<=1e5

思路：极其考验模板，反正我的spfa和zkw都挂了，就拿这题std做dijkstra费用流的板子了

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,ll>P;
 #define N  10000
 #define M  2100000
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=,inv2=(MOD+)/;
       double eps=1e-;
       int INF=0x7fffffff;;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};

 struct edge
 {
     int to,cap,cost,rev;
     edge(){}
     edge(int a,int b,int c,int d):
     to(a),cap(b),cost(c),rev(d){}
 };

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 struct MCMF
 {
     int V,h[N],dis[N],preV[N],preE[N];
     vector<edge> g[N];

     void init(int n)
     {
         V=n;
         rep(i,,V) g[i].clear();
     }

     void add(int u,int v,int cap,int cost)
     {
         g[u].pb(edge(v,cap,cost,g[v].size()));
         g[v].pb(edge(u,,-cost,g[u].size()-));
     }

     int minc_mf(int S,int T,int f,int &flow)
     {
         int res=;
         fill(h,h++V,);
         while(f)
         {
             priority_queue <PII,vector<PII>, greater<PII> > q;
             fill(dis,dis++V,INF);
             dis[S]=;
             q.push(MP(,S));
             while(!q.empty())
             {
                 PII now=q.top(); q.pop();
                 int u=now.se;
                 if(dis[u]<now.fi) continue;
                 for(int i=;i<g[u].size();i++)
                 {
                     edge e=g[u][i];
                     int v=e.to;
                     if(e.cap>&&dis[v]>dis[u]+e.cost+h[u]-h[v])
                     {
                         dis[v]=dis[u]+e.cost+h[u]-h[v];
                         preV[v]=u;
                         preE[v]=i;
                         q.push(MP(dis[v],v));
                     }

                 }
             }
             if(dis[T]==INF) break;
             rep(i,,V) h[i]+=dis[i];
             int t=f,k=T;
             while(k!=S)
             {
                 int e=preE[k];
                 t=min(t,g[preV[k]][preE[k]].cap);
                 k=preV[k];
             }
             f-=t; flow+=t; res+=t*h[T];
             k=T;
             while(k!=S)
             {
                 edge &e=g[preV[k]][preE[k]];
                 e.cap-=t;
                 g[k][e.rev].cap+=t;
                 k=preV[k];
             }
         }
         return res;
     }
 }mcmf;

 int num[N][],a[N],S,T,s,flow;

 int main()
 {
     int cas=read();
     s=;
     while(cas--)
     {
         int n=read(),k=read();
         rep(i,,n) a[i]=read();
         s=;
         rep(i,,n)
         {
             num[i][]=++s;
             num[i][]=++s;
         }
         S=++s; s++; T=++s;
         mcmf.init(s);
         mcmf.add(S,S+,k,);
         rep(i,,n) mcmf.add(S+,num[i][],,);
         rep(i,,n) mcmf.add(num[i][],num[i][],,-a[i]);
         rep(i,,n)
          rep(j,i+,n)
           if(a[j]>=a[i]) mcmf.add(num[i][],num[j][],,);
         rep(i,,n) mcmf.add(num[i][],T,,);
         flow=;
         int ans=-mcmf.minc_mf(S,T,INF,flow);
         printf("%d\n",ans);
     }
     return ;
15nb0 }

无敌zyd的优化建图，思路是每个i都只与后面有可能构成最优解的j连边，最坏情况下没有任何改进，但好像因为a[i]是随机数据跑的飞快

 #pragma GCC optimize("O3")
 #pragma G++ optimize("O3")
 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define rep(i,l,r) for(int i=l;i<=r;++i)
 #define per(i,l,r) for(int i=l;i>=r;--i)
 using namespace std;

 const int N = , M=;
 const int INF = 1e9;
 int n,m;
 int a[M];

 struct EG {
     int a, b, c, d, e;
 } eg[M];
 int head[N], en, S, T, SS;
 int flow, cost;
 int dis[N], pre[N], rem[N];
 bool inq[N];
 void insert(int u, int v, int w, int z) {
     eg[++en] = (EG) {v, head[u], w, z, u}; head[u] = en;
     eg[++en] = (EG) {u, head[v], , -z, v}; head[v] = en;
 }
 void Clear() {
     memset(head, , sizeof head);
     en = ;
 }
 bool Spfa() {
     for (int i = ; i <= SS; i++) dis[i] = INF, inq[i] = ;
     dis[S] = ;
     inq[S] = ;
     queue<int> Q; Q.push(S);
     pre[S] = , rem[S] = INF;
     while (!Q.empty()) {
         int u = Q.front(); Q.pop();
         inq[u] = ;
         for (int e = head[u]; e; e = eg[e].b) {
             int v = eg[e].a;
             if (eg[e].c >  && dis[u] + eg[e].d < dis[v]) {
                 dis[v] = dis[u] + eg[e].d;
                 pre[v] = e;
                 rem[v] = min(rem[u], eg[e].c);
                 if (!inq[v]) {
                     inq[v] = ;
                     Q.push(v);
                 }
             }
         }
     }
     if (dis[T] == INF) return ;
     flow += rem[T];
     cost += dis[T] * rem[T];
     int u = T;
     while (u != S) {
         eg[pre[u]].c -= rem[T];
         eg[pre[u] ^ ].c += rem[T];
         u = eg[pre[u]].e;
     }
     return ;
 }
 void MinCost() {
     flow = cost = ;
     while (Spfa());
 }

 int main() {
     //freopen("a.txt","r",stdin);

     int test_;
     cin>>test_;
     while (test_--) {
         scanf("%d%d",&n,&m);
         rep(i,,n) scanf("%d",a+i);

         Clear();
         S = n *  + , T = S + , SS = T + ;
         for (int i = ; i <= n; i++)
         {
             insert(i, i + n, , -a[i]);
             insert(i, i + n, m, );
             int mx = INF;
             for (int j = i + ; j <= n; j++)
             {
                 if (a[j] < mx && a[j] >= a[i])
                 {
                     insert(i + n, j, INF, );
                     mx = a[j];
                 }
             }
         }
         insert(S, SS, m, );
         for (int i = ; i <= n; i++)
         {
             insert(SS, i, , );
             insert(i + n, T, , );
         }
         MinCost();
         printf("%d\n", -cost);
     }
 }