LA4273 Post Offices

题目戳这里

村庄排序。状态\(f[j][i]\)表示考虑前\(i\)个村庄,造\(j\)个邮局且\(i\)造了邮局的最小代价。我们用\(Lb_i,Rb_i\)表示在第\(i\)个村庄造邮局,邮局最左和最右能够服务到的村庄。\(pre_i\)表示\(P\)的前缀和。

\[f[j][i] = C_i + \left\{ \begin{array}{ll} \min (f[j-1][k]) & Rb_k\ge Lb_i\\ \min (f[j-1][k]+pre_{Lb_i-1}-pre_{Rb_k}) & Rb_k < Lb_i \end{array} \right.\]

\(Rb_i\)来建线段树,分别维护\(f[j-1][k]\)\(f[j-1][k]-pre_{Rb_k}\)最小值即可。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
using namespace std;

const int maxn = 10010,maxm = 110; const int inf = 1<<30;
int N,M,X[maxn],C[maxn],R[maxn],Case,Lbound[maxn],Rbound[maxn],leaf[maxn];
int f[maxm][maxn],P[maxn],tree[4*maxn][2],ans;

inline int gi()
{
	char ch; int ret = 0,f = 1;
	do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-');
	if (ch == '-') f = -1,ch = getchar();
	do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9');
	return ret*f;
}

inline void build(int now,int l,int r,int i)
{
	if (l == r) { tree[now][0] = tree[now][1] = inf; leaf[l] = now; return; }
	int mid = (l+r) >> 1;
	build(now<<1,l,mid,i); build(now<<1|1,mid+1,r,i);
	tree[now][0] = tree[now][1] = inf;
}

inline void modify(int pos,int a,int b)
{
	if (a < tree[leaf[pos]][0])
	{
		tree[leaf[pos]][0] = a;
		for (int now = leaf[pos]>>1;now&&tree[now][0] > a;now >>= 1) tree[now][0] = a;
	}
	if (b < tree[leaf[pos]][1])
	{
		tree[leaf[pos]][1] = b;
		for (int now = leaf[pos]>>1;now&&tree[now][1] > b;now >>= 1) tree[now][1] = b;
	}
}

inline int query(int i,int now,int l,int r,int ql,int qr)
{
	if (l == ql&&r == qr) return tree[now][i];
	int mid = (l+r)>>1;
	if (qr <= mid) return query(i,now<<1,l,mid,ql,qr);
	else if (ql > mid) return query(i,now<<1|1,mid+1,r,ql,qr);
	else return min(query(i,now<<1,l,mid,ql,mid),query(i,now<<1|1,mid+1,r,mid+1,qr));
}

int main()
{
	freopen("4273.in","r",stdin);
	freopen("4273.out","w",stdout);
	while (++Case)
	{
		N = gi(); M = gi(); if (!N&&!M) break;
		printf("Case %d: ",Case);
		for (int i = 2;i <= N;++i) X[i] = gi();
		for (int i = 1;i <= N;++i) C[i] = gi();
		for (int i = 1;i <= N;++i) R[i] = gi();
		for (int i = 1;i <= N;++i) P[i] = gi()+P[i-1];
		for (int i = 1;i <= N;++i) Lbound[i] = lower_bound(X+1,X+N+1,X[i]-R[i])-X,Rbound[i] = upper_bound(X+1,X+N+1,X[i]+R[i])-X-1;
		ans = P[N];
		for (int i = 1;i <= N;++i) f[1][i] = C[i]+P[Lbound[i]-1],ans = min(ans,f[1][i]+P[N]-P[Rbound[i]]);
		for (int i = 2;i <= M;++i)
		{
			build(1,1,N,i-1);
			for (int j = i;j <= N;++j)
			{
				modify(Rbound[j-1],f[i-1][j-1],f[i-1][j-1]-P[Rbound[j-1]]);
				f[i][j] = query(0,1,1,N,Lbound[j],N);
				if (Lbound[j]-1) f[i][j] = min(f[i][j],query(1,1,1,N,1,Lbound[j]-1)+P[Lbound[j]-1]);
				f[i][j] += C[j];
				ans = min(ans,f[i][j]+P[N]-P[Rbound[j]]);
			}
		}
		cout << ans << endl;
	}	
	fclose(stdin); fclose(stdout);
	return 0;
}

posted @ 2017-02-15 23:22  lmxyy  阅读(254)  评论(0编辑  收藏  举报