bzoj 1520 [POI2006]Szk-Schools 费用流

[POI2006]Szk-Schools

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 743  Solved: 381
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Description

Input

Output

如果有可行解, 输出最小代价,否则输出NIE.

Sample Input

5
1 1 2 3
1 1 5 1
3 2 5 5
4 1 5 10
3 3 3 1

Sample Output

9

HINT

 

考虑数据范围应该就是比较裸的费用流吧

 

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstdlib>
  4 #include<algorithm>
  5 #include<cmath>
  6 #include<cstring>
  7 #include<set>
  8 #include<queue>
  9 #include<map>
 10 #define pa pair<int,int>
 11 #define mod 1000000007
 12 #define inf 1000000000
 13 #define ll long long
 14 using namespace std;
 15 int read()
 16 {
 17     int x=0,f=1;char ch=getchar();
 18     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 19     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 20     return x*f;
 21 }
 22 int n,cnt=1,T,tot,ans;
 23 int last[405],h[405],q[405],d[405];
 24 bool inq[405];
 25 struct data{int to,next,c,v;}e[100005];
 26 void ins(int u,int v,int w,int c)
 27 {
 28     e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;e[cnt].v=w;e[cnt].c=c;
 29 }
 30 void insert(int u,int v,int w,int c)
 31 {
 32     ins(u,v,w,c);
 33     ins(v,u,0,-c);
 34 }
 35 bool spfa()
 36 {
 37     memset(inq,0,sizeof(inq));
 38     int head=0,tail=1;
 39     for(int i=0;i<=T;i++)d[i]=inf;
 40     q[0]=T;d[T]=0;inq[T]=1;
 41     while(head!=tail)
 42     {
 43         int now=q[head];head++;if(head==405)head=0;
 44         for(int i=last[now];i;i=e[i].next)
 45             if(e[i^1].v&&d[now]-e[i].c<d[e[i].to])
 46             {
 47                 d[e[i].to]=d[now]-e[i].c;
 48                 if(!inq[e[i].to])
 49                 {
 50                     inq[e[i].to]=1;
 51                     q[tail]=e[i].to;
 52                     tail++;if(tail==405)tail=0;
 53                 }
 54             }
 55         inq[now]=0;
 56     }
 57     return d[0]!=inf;
 58 }
 59 int dfs(int x,int f)
 60 {
 61     inq[x]=1;
 62     if(x==T)return f;
 63     int used=0,w;
 64     for(int i=last[x];i;i=e[i].next)
 65         if(!inq[e[i].to]&&e[i].v&&d[x]-e[i].c==d[e[i].to])
 66         {
 67             w=f-used;
 68             w=dfs(e[i].to,min(e[i].v,w));
 69             ans+=w*e[i].c;
 70             e[i].v-=w;e[i^1].v+=w;
 71             used+=w;if(used==f)return f;
 72         }
 73     return used;
 74 }
 75 void zkw()
 76 {
 77     while(spfa())
 78     {
 79         inq[T]=1;
 80         while(inq[T])
 81         {
 82             memset(inq,0,sizeof(inq));
 83             tot+=dfs(0,inf);
 84         }
 85     }
 86 }
 87 int main()
 88 {
 89     n=read();T=2*n+1;
 90     for(int i=1;i<=n;i++)insert(0,i,1,0);
 91     for(int i=1;i<=n;i++)insert(i+n,T,1,0);
 92     for(int i=1;i<=n;i++)
 93     {
 94         int m=read(),a=read(),b=read(),k=read();
 95         for(int j=a;j<=b;j++)
 96             insert(i,n+j,1,abs((j-m)*k));
 97     }
 98     zkw();
 99     if(tot!=n)puts("NIE");
100     else printf("%d\n",ans);
101     return 0;
102 }

 

posted @ 2018-04-15 14:04  Kaiser-  阅读(115)  评论(0)  编辑  收藏  举报