[BZOJ1672][Usaco2005 Dec]Cleaning Shifts 清理牛棚 线段树优化DP

链接

题意：给你一些区间，每个区间都有一个花费，求覆盖区间 \([S,T]\) 的最小花费

题解

先将区间排序

设 \(f[i]\) 表示决策到第 \(i\) 个区间，覆盖满 \(S\dots R[i]\) 的最小花费

显然 \(f[i]=\min_{R[j]\ge L[i]}f[j]+w[i]\)

按照区间建线段树，插右端点即可

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i(a);i<=(b);++i)
using namespace std;
typedef long long ll;
inline int read(){char c;int w;
	while(!isdigit(c=getchar()));w=c&15;
	while(isdigit(c=getchar()))w=w*10+(c&15);return w;
}
inline char smax(int&x,const int&y){return x<y?x=y,1:0;}
inline char smin(ll&x,const ll&y){return x>y?x=y,1:0;}
const int N=100005;
struct data{int l,r,w;}a[N];
inline bool cmp(data x,data y){return x.r<y.r;}
int n,s,t;
ll minv[N<<2],f[N];
#define ls o<<1
#define rs o<<1|1
inline void ins(int o,int l,int r,int x,ll w){
	smin(minv[o],w);if(l==r)return;int mid=l+r>>1;
	x<=mid?ins(ls,l,mid,x,w):ins(rs,mid+1,r,x,w);
}
inline ll query(int o,int l,int r,int x,int y){
	if(x>r||y<l)return minv[0];
	if(x<=l&&r<=y)return minv[o];
	int mid=l+r>>1;
	return min(query(ls,l,mid,x,y),query(rs,mid+1,r,x,y));
}
int main(){
	n=read(),s=read()+1,t=read()+1;
	REP(i,1,n)a[i]=(data){read()+1,read()+1,read()};
	sort(a+1,a+1+n,cmp);
	memset(f,0x3f,sizeof f);memset(minv,0x3f,sizeof minv);
	#define all 1,s-1,t
	ins(all,s-1,0);f[s-1]=0;
	REP(i,1,n)if(a[i].r>=s&&a[i].l<=t){
		f[i]=query(all,max(s-1,a[i].l-1),min(a[i].r-1,t))+a[i].w;
		ins(all,min(a[i].r,t),f[i]);
	}
	ll ans=minv[0];
	for(int i=n;i;--i)if(a[i].r>=t){
		smin(ans,f[i]);
	}else break;
	if(ans==minv[0])puts("-1");else cout<<ans;
	return 0;
}