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。

 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\n",-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 }

 

posted @ 2016-06-24 19:59  GFY  阅读(228)  评论(0)  编辑  收藏  举报