Description

You are planning to build housing on a street. There are n spots available on the street on which you can build a house. The spots are labeled from 1 to n from left to right. In each spot, you can build a house with an integer height between 0 and h.

In each spot, if a house has height a, you can gain a2 dollars from it.

The city has m zoning restrictions though. The i-th restriction says that if the tallest house from spots li to ri is strictly more than xi, you must pay a fine of ci.

You would like to build houses to maximize your profit (sum of dollars gained minus fines). Determine the maximum profit possible.

Input

The first line contains three integers n,h,m (1≤n,h,m≤50) — the number of spots, the maximum height, and the number of restrictions, respectively.

Each of the next m lines contains four integers li,ri,xi,ci (1≤li≤ri≤n, 0≤xi≤h, 1≤ci≤5000).

Output

Print a single integer denoting the maximum profit you can make.

Solution

大意就是建房子，房子最大高度为h，对于高度为i的建筑，收益为i*i。现在有m个限制，对于第i个限制，在li~ri这段区间内，高度如果有超过xi的需要罚款ci元。问最大收益

考虑最小割来求损失

首先设一个超级源S和超级汇T

将每一个点拆成0~h-1即h个点

将0点与S连一条正无穷的边

随后对于这h个点：

第i个点与第i+1个点之间连一条hh-ii的边，表示损失

割掉第i个点与第i+1个点之间的连边就代表选了第i个点

随后，对于每个限制

li~ri区间内的第xi个点与该点连一条正无穷的边，因为这条边不可能被割

该点再与T连一条ci的边，表示损失

最后输出即可

(第一次打sap，之前都在打dinic呢）

#include <cstdio>

#include <algorithm>

#define inf 1000000007

#define S 4000

#define T 4001

#define M 20001

#define N 5001

#define open(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);

#define id(x,y) (x-1)*(h+1)+y+1

using namespace std;

int len,n,h,m,l,r,x,c,cnt,ans,i,j,go[M],to[M],last[N],w[M],dis[N],gap[N];

void make(int x,int y,int z)

{

    go[++len]=y;to[len]=last[x];w[len]=z;last[x]=len;

}

void add(int x,int y,int z)

{

    make(x,y,z);

    make(y,x,0);

}

int sap(int x,int flow)

{

    if (x==T) return flow;

    int tmp,have=flow;

    for (int i=last[x];i;i=to[i])

    {

        if (w[i] && dis[x]==dis[go[i]]+1)

        {

            tmp=sap(go[i],min(have,w[i]));

            w[i]-=tmp;w[i^1]+=tmp;have-=tmp;

            if (!have) return flow;

        }

    }

    if (!--gap[dis[x]]) dis[S]=T;

    ++gap[++dis[x]];

    return flow-have;

}

int main()

{

    open("restrictions");

    scanf("%d%d%d",&n,&h,&m);

    len=1;

    for (i=1;i<=n;i++)

    {

        add(S,id(i,0),inf);

        for (j=0;j<h;j++)

        {

            add(id(i,j),id(i,j+1),h*h-j*j);

        }

    }

    cnt=id(n,h);

    for (i=1;i<=m;i++)

    {

        scanf("%d%d%d%d",&l,&r,&x,&c);

        if (x==h) continue;

        add(++cnt,T,c);

        for (j=l;j<=r;j++)

            add(id(j,x+1),cnt,inf);

    }

    ans=n*h*h;

    while (dis[S]<T)

		ans-=sap(S,inf);

    printf("%d",ans);

    return 0;

}

巴特西

Codeforces1146G. Zoning Restrictions

Description

Input

Output

Solution

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