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。
1 #include<cstdio> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstring> 6 #define inf 0x7fffffff 7 int tot,go[200005],next[200005],first[200005],flow[200005],cost[200005]; 8 int dis[200005],vis[200005],op[200005],edge[200005],from[200005],ans,a[200005]; 9 int n,K,S,T,c[200005]; 10 int read(){ 11 int t=0,f=1;char ch=getchar(); 12 while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();} 13 while ('0'<=ch&&ch<='9'){t=t*10+ch-'0';ch=getchar();} 14 return t*f; 15 } 16 void insert(int x,int y,int z,int l){ 17 tot++; 18 go[tot]=y; 19 next[tot]=first[x]; 20 first[x]=tot; 21 flow[tot]=z; 22 cost[tot]=l; 23 } 24 void add(int x,int y,int z,int l){ 25 insert(x,y,z,l);op[tot]=tot+1; 26 insert(y,x,0,-l);op[tot]=tot-1; 27 } 28 bool spfa(){ 29 for (int i=S;i<=T;i++) dis[i]=0x3f3f3f3f,vis[i]=0; 30 int h=1,t=1;vis[S]=1;c[1]=S;dis[S]=0; 31 while (h<=t){ 32 int now=c[h++]; 33 for (int i=first[now];i;i=next[i]){ 34 int pur=go[i]; 35 if (dis[pur]>dis[now]+cost[i]&&flow[i]){ 36 dis[pur]=dis[now]+cost[i]; 37 from[pur]=now; 38 edge[pur]=i; 39 if (vis[pur]) continue; 40 c[++t]=pur; 41 vis[pur]=1; 42 } 43 } 44 vis[now]=0; 45 } 46 return dis[T]!=0x3f3f3f3f; 47 } 48 void updata(){ 49 int mn=0x7fffffff; 50 for (int i=T;i!=S;i=from[i]){ 51 mn=std::min(mn,flow[edge[i]]); 52 } 53 for (int i=T;i!=S;i=from[i]){ 54 flow[edge[i]]-=mn; 55 flow[op[edge[i]]]+=mn; 56 ans+=mn*cost[edge[i]]; 57 } 58 } 59 void mncostflow(){ 60 ans=0; 61 while (spfa()) updata(); 62 printf("%d ",-ans); 63 } 64 int main(){ 65 n=read();K=read();S=0;T=2*n+3; 66 for (int i=1;i<=3*n;i++) 67 a[i]=read(); 68 for (int i=1;i<=2*n+1;i++) 69 add(i+1,i,inf,0); 70 add(1,T,K,0); 71 add(S,2*n+2,K,0); 72 for (int i=1;i<=n;i++) 73 add(i+1,1,1,-a[i]); 74 for (int i=n+1;i<=n+n;i++) 75 add(i+1,i+1-n,1,-a[i]); 76 for (int i=n+n+1;i<=n*3;i++) 77 add(2*n+2,i-n+1,1,-a[i]); 78 mncostflow(); 79 }