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=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,cnt=1,T,tot,ans;
int last[405],h[405],q[405],d[405];
bool inq[405];
struct data{int to,next,c,v;}e[100005];
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,0,-c);
}
bool spfa()
{
    memset(inq,0,sizeof(inq));
    int head=0,tail=1;
    for(int i=0;i<=T;i++)d[i]=inf;
    q[0]=T;d[T]=0;inq[T]=1;
    while(head!=tail)
    {
        int now=q[head];head++;if(head==405)head=0;
        for(int i=last[now];i;i=e[i].next)
            if(e[i^1].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]=1;
                    q[tail]=e[i].to;
                    tail++;if(tail==405)tail=0;
                }
            }
        inq[now]=0;
    }
    return d[0]!=inf;
}
int dfs(int x,int f)
{
    inq[x]=1;
    if(x==T)return f;
    int used=0,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^1].v+=w;
            used+=w;if(used==f)return f;
        }
    return used;
}
void zkw()
{
    while(spfa())
    {
        inq[T]=1;
        while(inq[T])
        {
            memset(inq,0,sizeof(inq));
            tot+=dfs(0,inf);
        }
    }
}
int main()
{
    n=read();T=2*n+1;
    for(int i=1;i<=n;i++)insert(0,i,1,0);
    for(int i=1;i<=n;i++)insert(i+n,T,1,0);
    for(int i=1;i<=n;i++)
    {
        int m=read(),a=read(),b=read(),k=read();
        for(int j=a;j<=b;j++)
            insert(i,n+j,1,abs((j-m)*k));
    }
    zkw();
    if(tot!=n)puts("NIE");
    else printf("%d
",ans);
    return 0;
}