hdu3669

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链接:hdu3669

题解

  • 把给的矩形以长为第一关键词宽为第二关键词降序排列,去掉包含的矩形,定义dp[j][i]表示前i个矩形挖了j个洞的最小花费

则dp[j][i]=min(dp[j-1][k-1]+x[k]*y[i])

依然考虑p<k,且k比p优则

即

  • 维护一个上凸曲线即可.(注意细节,并且如果挖洞数增加了花费没有变小那么break,不然会T)
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
#define xx(i) a[i].first
#define yy(i) a[i].second
using namespace std;
const int N=1e5+7;
typedef pair<int,int> s;
s b[N],a[N];
ll dp[107][N];
int n,nn,m,q[N];
inline int cmp(s x,s y){return x.first==y.first?x.second>y.second:x.first>y.first;}
inline ll Y(int x,int y,int c){return dp[c][x]-dp[c][y];}
inline ll X(int x,int y){return 1LL*xx(y+1)-xx(x+1);}
int main(){
	while(~scanf("%d%d",&nn,&m)){
		int l=1,r=1;n=0;
		rep(i,1,nn)scanf("%d%d",&b[i].first,&b[i].second);
		sort(b+1,b+nn+1,cmp);
		while(l<=nn){
			r=l+1;
			while(r<=nn&&b[l].second>=b[r].second)
				++r;
			a[++n]=b[l];
			l=r;
		}
		rep(i,1,n)dp[1][i]=1LL*xx(1)*yy(i);
		ll ans=dp[1][n];
		rep(j,2,m){
		int head=1,tail=1;
			rep(i,1,n){
				while(head<tail&&Y(q[head+1],q[head],j-1)<=X(q[head+1],q[head])*yy(i))++head;
				dp[j][i]=dp[j-1][q[head]]+1LL*xx(q[head]+1)*yy(i);
				while(head<tail&&Y(q[tail],i,j-1)*X(q[tail-1],q[tail])<=Y(q[tail-1],q[tail],j-1)*X(q[tail],i))--tail;
				q[++tail]=i;
			}
			if(ans<=dp[j][n])break;
			else ans=dp[j][n];
		}
		printf("%lld\n",ans);
	}
	return 0;
}