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

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<set>
 #include<queue>
 #include<map>
 #define pa pair<int,int>
 #define mod 1000000007
 #define inf 1000000000
 #define ll long long
 using namespace std;
 int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,cnt=,T,tot,ans;
 int last[],h[],q[],d[];
 bool inq[];
 struct data{int to,next,c,v;}e[];
 void ins(int u,int v,int w,int c)
 {
     e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;e[cnt].v=w;e[cnt].c=c;
 }
 void insert(int u,int v,int w,int c)
 {
     ins(u,v,w,c);
     ins(v,u,,-c);
 }
 bool spfa()
 {
     memset(inq,,sizeof(inq));
     int head=,tail=;
     for(int i=;i<=T;i++)d[i]=inf;
     q[]=T;d[T]=;inq[T]=;
     while(head!=tail)
     {
         int now=q[head];head++;if(head==)head=;
         for(int i=last[now];i;i=e[i].next)
             if(e[i^].v&&d[now]-e[i].c<d[e[i].to])
             {
                 d[e[i].to]=d[now]-e[i].c;
                 if(!inq[e[i].to])
                 {
                     inq[e[i].to]=;
                     q[tail]=e[i].to;
                     tail++;if(tail==)tail=;
                 }
             }
         inq[now]=;
     }
     return d[]!=inf;
 }
 int dfs(int x,int f)
 {
     inq[x]=;
     if(x==T)return f;
     int used=,w;
     for(int i=last[x];i;i=e[i].next)
         if(!inq[e[i].to]&&e[i].v&&d[x]-e[i].c==d[e[i].to])
         {
             w=f-used;
             w=dfs(e[i].to,min(e[i].v,w));
             ans+=w*e[i].c;
             e[i].v-=w;e[i^].v+=w;
             used+=w;if(used==f)return f;
         }
     return used;
 }
 void zkw()
 {
     while(spfa())
     {
         inq[T]=;
         while(inq[T])
         {
             memset(inq,,sizeof(inq));
             tot+=dfs(,inf);
         }
     }
 }
 int main()
 {
     n=read();T=*n+;
     for(int i=;i<=n;i++)insert(,i,,);
     for(int i=;i<=n;i++)insert(i+n,T,,);
     for(int i=;i<=n;i++)
     {
         int m=read(),a=read(),b=read(),k=read();
         for(int j=a;j<=b;j++)
             insert(i,n+j,,abs((j-m)*k));
     }
     zkw();
     if(tot!=n)puts("NIE");
     else printf("%d\n",ans);
     return ;
 }