BZOJ1520: [POI2006]Szk-Schools

1520: [POI2006]Szk-Schools

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 342  Solved: 173
[Submit][Status]

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

Source

题解：

裸的最小费用最大流。。。

代码：

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define inf 1000000000

#define maxn 500+100

#define maxm 250000

#define eps 1e-10

#define ll long long

#define pa pair<int,int>

#define for0(i,n) for(int i=0;i<=(n);i++)

#define for1(i,n) for(int i=1;i<=(n);i++)

#define for2(i,x,y) for(int i=(x);i<=(y);i++)

#define for3(i,x,y) for(int i=(x);i>=(y);i--)

#define mod 1000000007

using namespace std;

inline 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=10*x+ch-'0';ch=getchar();}

    return x*f;

}
int n,m,k,mincost,maxflow,tot=1,s,t,head[maxn],q[maxm],d[maxn],from[2*maxm];

bool v[maxn];

struct edge{int from,next,go,v,c;}e[2*maxm];

void ins(int x,int y,int z,int w)

{

    e[++tot].go=y;e[tot].from=x;e[tot].v=z;e[tot].c=w;e[tot].next=head[x];head[x]=tot;

}

void insert(int x,int y,int z,int w)

{

    ins(x,y,z,w);ins(y,x,0,-w);

}

bool spfa()

{

    for (int i=s;i<=t;i++){v[i]=0;d[i]=inf;}

    int l=0,r=1,y;q[1]=s;d[s]=0;v[0]=1;

    while(l!=r)

    {

        int x=q[++l];if(l==maxn)l=0;v[x]=0;

        for (int i=head[x];i;i=e[i].next)

         if(e[i].v&&d[x]+e[i].c<d[y=e[i].go])

         {

            d[y]=d[x]+e[i].c;from[y]=i;

            if(!v[y]){v[y]=1;q[++r]=y;if(r==maxn)r=0;}

         }

    }

    return d[t]!=inf;

}

void mcf()

{

    while(spfa())

    {

        int tmp=inf;

        for(int i=from[t];i;i=from[e[i].from]) tmp=min(tmp,e[i].v);

        mincost+=d[t]*tmp;maxflow+=tmp;

        for(int i=from[t];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;}

    }

}

int main()

{

    freopen("input2.txt","r",stdin);

    freopen("output3.txt","w",stdout);

    n=read();s=0;t=2*n+1;
    for1(i,n)
    {
        insert(s,i,1,0);insert(i+n,t,1,0);
        int x=read(),l=read(),r=read(),y=read();
        for2(j,l,r)insert(i,j+n,1,y*(abs(j-x)));
    }
    mcf();
    if(maxflow==n)printf("%d
",mincost);else puts("NIE");

    return 0;

}