Cleaning Shifts POJ - 3171 （线段树优化dp）

Farmer John’s cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn.

Farmer John has N (1 <= N <= 10,000) cows who are willing to do some cleaning. Because dust falls continuously, the cows require that the farm be continuously cleaned during the workday, which runs from second number M to second number E during the day (0 <= M <= E <= 86,399). Note that the total number of seconds during which cleaning is to take place is E-M+1. During any given second M..E, at least one cow must be cleaning.

Each cow has submitted a job application indicating her willingness to work during a certain interval T1..T2 (where M <= T1 <= T2 <= E) for a certain salary of S (where 0 <= S <= 500,000). Note that a cow who indicated the interval 10..20 would work for 11 seconds, not 10. Farmer John must either accept or reject each individual application; he may NOT ask a cow to work only a fraction of the time it indicated and receive a corresponding fraction of the salary.

Find a schedule in which every second of the workday is covered by at least one cow and which minimizes the total salary that goes to the cows.
Input
Line 1: Three space-separated integers: N, M, and E.

Lines 2..N+1: Line i+1 describes cow i’s schedule with three space-separated integers: T1, T2, and S.
Output
Line 1: a single integer that is either the minimum total salary to get the barn cleaned or else -1 if it is impossible to clean the barn.
Sample Input
3 0 4
0 2 3
3 4 2
0 0 1
Sample Output
5
Hint
Explanation of the sample:

FJ has three cows, and the barn needs to be cleaned from second 0 to second 4. The first cow is willing to work during seconds 0, 1, and 2 for a total salary of 3, etc.

Farmer John can hire the first two cows.

今天有点难过。难过。
代码有点问题，明天调一下。
4.16更新：
思路：首先把和目标区间无关的区间去除，然后以右区间为标准从小到大排序，然后依次考虑每个区间的所带来的状态转移。
设$f[i]$为覆盖m到i的最小花费。那么当考虑当前的区间时设为$[a_i,b_i]$,且花费为$c_i$，则

#include<iostream>
#include<cstdio>
#include<stack>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cmath>
#define maxx 100005
using namespace std;
int n,m,e;
const long long INF=1000000000000l;
struct node
{
    int l,r;
    int c;
}p[10005];
bool cmp(node x1,node x2)
{
    return x1.r<x2.r;
}
long long c[90000<<2];
void build(int st,int l,int r)
{
    c[st]=INF;
    if(l==r)
        return ;
    int mid=(l+r)>>1;
    build(st<<1,l,mid);
    build(st<<1|1,mid+1,r);
}
void update(int st,int l,int r,int pos,long long x)
{
    if(l==r)
    {
        c[st]=x;
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid)
        update(st<<1,l,mid,pos,x);
    else
        update(st<<1|1,mid+1,r,pos,x);
    c[st]=min(c[st<<1],c[st<<1|1]);
}
long long query(int st,int l,int r,int L,int R)
{
    if(L<=l&&r<=R)
        return c[st];
    int mid=(l+r)>>1;
    long long ans=INF;
    if(mid>=L)
         ans=min(ans,query(st<<1,l,mid,L,R));
    if(mid<R)
        ans=min(ans,query(st<<1|1,mid+1,r,L,R));
    return ans;
}
int main()
{
    cin>>n>>m>>e;
    int x,y,w;
    int cnt=0;
    int up=0;
    int maxL=1000000000;
    int maxR=-1;
    for(int i=0;i<n;i++)
    {
        scanf("%d%d%d",&x,&y,&w);
        if(y<m||e<x)//去除无关的区间
            continue;
        maxL=min(maxL,x);
        maxR=max(maxR,y);
        p[cnt].l=x;
        p[cnt].r=y;
        p[cnt++].c=w;
    }
    maxL--;//这个需要注意的。
    build(1,maxL,maxR);//建树
    update(1,maxL,maxR,m-1,0);

    sort(p,p+cnt,cmp);
    for(int i=0;i<cnt;i++)
    {
        int l=p[i].l;
        int r=p[i].r;
        long long temp=query(1,maxL,maxR,r,r);
        long long ans=query(1,maxL,maxR,l-1,r-1)+p[i].c;
        if(temp>ans)
            update(1,maxL,maxR,r,ans);
    }//状态转移
    long long ans=query(1,maxL,maxR,e,maxR);
    if(ans==INF)
        cout<<-1<<endl;
    else
        cout<<ans<<endl;
    return 0;
}