【刷题】BZOJ 4254 Aerial Tramway

Description

You own a park located on a mountain, which can be described as a sequence of n points (xi, yi) from left to right, where xi,yi>0, xi<xi+1, yi!=yi+1 (that means there will not be horizontal segments in the mountain skyline), illustrated below(the numbers are the corresponding x-coordinate): Since the mountain is very sloppy, some aerial tramways across the park would be very helpful. In the figure above, people can go from p4 to p9 directly, by taking a tram. Otherwise he must follow a rather

zigzag path: p4-p5-p6-p7-p8-p9.

Your job is to design an aerial tramway system. There should be exactly m trams, each following a horizontal segment in the air, between two points pi and pj. "Horizontal" means yi=yj, “in the air" means all the points in between are strictly below, i.e. yk<yi for every i<k<j. For example, no tram can travel between p2 and p9, because p4 is not strictly below p2-p9. However, you can have two trams, one

from p2 to p4, and one p4 to p9. There is another important restriction: no point can be strictly below k or more tramways, because it’ll be dangerous. For example, if k=3, we cannot build these 3 tramways simultaneously: p1-p14, p4-p9, p6-p8, because p7 would be dangerous. You want to make this system as useful as possible, so you would like to maximize the total length of all tramways. For example, if m=3, k=3, the best design for the figure above is p1-p14, p2-p4 and p4-p9,the total length is 20. If m=3, k=2, you have to replace p1-p14 with p11-p13, the total length becomes 9.

Input

There will be at most 200 test cases. Each case begins with three integers n, m and k(1<=n,m<=200, 2<=k<=10), the number of points, the number of trams in your design and the dangerous parameter introduced earlier. The next line contains n pairs of positive integers xi and yi.(1<=xi,yi<=10^5).

Output

For each test case, print the case number and the maximal sum. If it is impossible to have exactly m tramways, print -1.

Sample Input

14 3 3

1 8

2 6

3 4

4 6

5 3

6 4

7 1

8 4

9 6

10 4

11 6

12 5

13 6

14 8

14 3 2

1 8

2 6

3 4

4 6

5 3

6 4

7 1

8 4

9 6

10 4

11 6

12 5

13 6

14 8

Sample Output

Case 1: 20

Case 2: 9

HINT

2015年湖南省大学生程序设计竞赛

Solution

我们把所有可以建的索道当做一个个区间

会发现这些区间之间要么互不相关，要么互相包含

于是就可以将这些区间建成一棵树，我们要做的就是在这棵树里选择正好 $m$ 个点，使得点权和最大，并且对于每一条深度递增的链，链上选的点必须小于 $k$ 个

这个树形dp一下就好了，设 $f_{i,j,p}$ 代表到树上 $i$ 点，其子树中选 $j$ 个，深度递增的链中选的最多的链选的个数为 $k$

转移方程为 $f_{u,j+r,max(p,t)}=\max\{f_{u,j,p}+f_{v,r,t}\}$

用前缀最大值省去最后第三维的枚举

就可以过了

#include<bits/stdc++.h>

#define ui unsigned int

#define ll long long

#define db double

#define ld long double

#define ull unsigned long long

const int MAXN=200+10,MAXP=100000+10,MAXK=20+10,inf=0x3f3f3f3f;

int n,m,k,e,beg[MAXN],nex[MAXN<<1],to[MAXN<<1],cnt,fa[MAXN],val[MAXN],f[MAXN][MAXN][MAXK],g[MAXN][MAXK],size[MAXN],ans;

struct node{

	int id,x,y;

	inline bool operator < (const node &A) const {

		return x<A.x;

	};

};

node mountain[MAXN];

std::vector<node> V[MAXP];

struct interval{

	int l,r;

	inline bool operator < (const interval &A) const {

		return r-l>A.r-A.l;

	};

};

interval line[MAXN];

template<typename T> inline void read(T &x)

{

	T data=0,w=1;

	char ch=0;

	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();

	if(ch=='-')w=-1,ch=getchar();

	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();

	x=data*w;

}

template<typename T> inline void write(T x,char ch='\0')

{

	if(x<0)putchar('-'),x=-x;

	if(x>9)write(x/10);

	putchar(x%10+'0');

	if(ch!='\0')putchar(ch);

}

template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}

template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}

template<typename T> inline T min(T x,T y){return x<y?x:y;}

template<typename T> inline T max(T x,T y){return x>y?x:y;}

inline void insert(int x,int y)

{

	to[++e]=y;

	nex[e]=beg[x];

	beg[x]=e;

}

inline void dfs(int x)

{

	for(register int i=0;i<=m;++i)

		for(register int j=0;j<k;++j)f[x][i][j]=-inf;

	f[x][0][0]=0;

	for(register int i=beg[x];i;i=nex[i])

		if(to[i]==fa[x])continue;

		else

		{

			dfs(to[i]);

			static int q1,q2;

			for(register int j=0;j<=size[x]+size[to[i]];++j)

				for(register int p=0;p<k;++p)g[j][p]=f[x][j][p];

			for(register int j=0;j<=size[x];++j)

				for(register int p=0;p<=min(size[to[i]],m-j);++p)

				{

					q1=-inf,q2=-inf;

					for(register int t=0;t<k;++t)

					{

						chkmax(q1,f[x][j][t]),chkmax(q2,f[to[i]][p][t]);

						chkmax(g[j+p][t],f[x][j][t]+q2);

						chkmax(g[j+p][t],f[to[i]][p][t]+q1);

					}

				}

			size[x]+=size[to[i]];

			for(register int j=0;j<=min(m,size[x]);++j)

				for(register int p=0;p<k;++p)f[x][j][p]=g[j][p];

		}

	if(x)

	{

		size[x]++;

		for(register int i=min(size[x],m)-1;i>=0;--i)

			for(register int j=k-2;j>=0;--j)chkmax(f[x][i+1][j+1],f[x][i][j]+val[x]);

	}

}

int main()

{

	static int cases=0;

	while(scanf("%d%d%d",&n,&m,&k)!=EOF)

	{

		for(register int i=1;i<=1e5;++i)V[i].clear();

		e=0;memset(beg,0,sizeof(beg));cnt=0;ans=-1;

		memset(size,0,sizeof(size));

		memset(fa,0,sizeof(fa));

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

		{

			int x,y;read(x);read(y);

			mountain[i]=(node){i,x,y};

		}

		std::sort(mountain+1,mountain+n+1);

		for(register int i=1;i<=n;++i)V[mountain[i].y].push_back(mountain[i]);

		for(register int i=1;i<=1e5;++i)

			for(register int j=1,lt=V[i].size();j<lt;++j)

			{

				int l=V[i][j-1].id+1,r=V[i][j].id-1,nowh=0;

				for(register int k=l;k<=r;++k)chkmax(nowh,mountain[k].y);

				if(nowh<i)line[++cnt]=(interval){V[i][j-1].x,V[i][j].x};

			}

		std::sort(line+1,line+cnt+1);

		for(register int i=1;i<=cnt;++i)

			for(register int j=i-1;j>=1;--j)

				if(line[j].l<=line[i].l&&line[i].r<=line[j].r)

				{

					insert(i,j),insert(j,i);

					fa[i]=j;break;

				}

		for(register int i=1;i<=cnt;++i)

		{

			val[i]=line[i].r-line[i].l;

			if(!fa[i])insert(0,i),insert(i,0);

		}

		fa[0]=-1;dfs(0);

		for(register int i=0;i<k;++i)chkmax(ans,f[0][m][i]);

		printf("Case %d: %d\n",++cases,ans);

	}

	return 0;

}