BZOJ 3550 Vacation

http://www.lydsy.com/JudgeOnline/problem.php?id=3550

题意:有3N个数，你需要选出一些数，首先保证任意长度为N的区间中选出的数的个数<=K个，其次要保证选出的数的个数最大。

思路:和这题类似http://www.cnblogs.com/qzqzgfy/p/5612261.html

可以转换成不等式然后求费用流。

 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #define inf 0x7fffffff
 int tot,go[],next[],first[],flow[],cost[];
 int dis[],vis[],op[],edge[],from[],ans,a[];
 int n,K,S,T,c[];
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void insert(int x,int y,int z,int l){
    tot++;
    go[tot]=y;
    next[tot]=first[x];
    first[x]=tot;
    flow[tot]=z;
    cost[tot]=l;
 }
 void add(int x,int y,int z,int l){
    insert(x,y,z,l);op[tot]=tot+;
    insert(y,x,,-l);op[tot]=tot-;
 }
 bool spfa(){
    for (int i=S;i<=T;i++) dis[i]=0x3f3f3f3f,vis[i]=;
    int h=,t=;vis[S]=;c[]=S;dis[S]=;
    while (h<=t){
        int now=c[h++];
        for (int i=first[now];i;i=next[i]){
              int pur=go[i];
              if (dis[pur]>dis[now]+cost[i]&&flow[i]){
                 dis[pur]=dis[now]+cost[i];
                 from[pur]=now;
                 edge[pur]=i;
                 if (vis[pur]) continue;
                 c[++t]=pur;
                 vis[pur]=;
             }
         }
         vis[now]=;
     }
     return dis[T]!=0x3f3f3f3f;
 }
 void updata(){
     int mn=0x7fffffff;
     for (int i=T;i!=S;i=from[i]){
           mn=std::min(mn,flow[edge[i]]);
     }
     for (int i=T;i!=S;i=from[i]){
           flow[edge[i]]-=mn;
           flow[op[edge[i]]]+=mn;
           ans+=mn*cost[edge[i]];
     }
 }
 void mncostflow(){
    ans=;
    while (spfa()) updata();
    printf("%d\n",-ans);
 }
 int main(){
     n=read();K=read();S=;T=*n+;
     for (int i=;i<=*n;i++)
         a[i]=read();
     for (int i=;i<=*n+;i++)
         add(i+,i,inf,);
     add(,T,K,);
     add(S,*n+,K,);
     for (int i=;i<=n;i++)
         add(i+,,,-a[i]);
     for (int i=n+;i<=n+n;i++)
         add(i+,i+-n,,-a[i]);
     for (int i=n+n+;i<=n*;i++)
         add(*n+,i-n+,,-a[i]);
     mncostflow();
 }