POJ 3171.Cleaning Shifts-区间覆盖最小花费-dp+线段树优化(单点更新、区间查询最值)

Cleaning Shifts

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4721		Accepted: 1593

Description

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.

Source

USACO 2005 December Silver

题意就是给你一些小区间，用小区间去覆盖大区间，求最小花费。

数据结构优化动态规划，这是一个区间最值问题。

将小区间按照右端点排序，然后遍历小区间，每次都找小区间左边到小区间右边的最小值，然后最小值更新，只单点更新右端点就可以(降维)。

一开始想的区间更新，按照贴纸，直接成段更新，后来发现思路是错的，因为区间更新会覆盖掉其他的值。所以要用单点更新。

线段树每个点表示到当前区间段的最优解，直接初始化为inf，然后M-1更新为0，就可以。

每次都是查找a[i].l-1到a[i].r的最小值(因为是片段，所以从a[i].l-1开始查找)，然后更新a[i].r。直接线段树维护区间的最小值。

动态规划方程式就是dp[i]=min(query(a[i].l-1,a[i].r))+a[i].val;表示覆盖到a[i].r的最小花费。因为小区间不能间断，所以要这样查询更新。

因为直接遍历的，所以dp[i]这个方程式都可以省略，直接查值然后更新值到线段树里就可以了。

因为每次都是从a[i].l-1开始，而且题目数据最小可能为0，所以直接端点都+2，这样，最小的值查找的时候也是从1开始的。

写题的时候，不能乱秀，容易把自己秀死。。。

自己写的时候，特判了一下如果最左端点和最端点不能包含大区间，直接-1输出。结果忘了我是按照右端点排序的，右边是对的，左边的不一定是对的。。。

最后直接查询大区间右端点和小区间最右端点。判断一下值就可以了。

如果ans<inf说明都能覆盖到，如果有inf说明有的地方没被覆盖到。

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const ll inf=1e18;
const int maxn=1e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

struct node{
    int l,r;
    ll val;

    friend bool operator<(node a,node b){
        if(a.r==b.r) return a.l<b.l;
        return a.r<b.r;
    }

}a[maxn];

ll tree[maxn<<2];

void pushup(int rt)
{
    tree[rt]=min(tree[rt<<1],tree[rt<<1|1]);
}

void build(int l,int r,int rt)
{
    if(l==r){
        tree[rt]=inf;
        return ;
    }

    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int pos,ll c,int l,int r,int rt)
{
    if(l==r){
        tree[rt]=min(tree[rt],c);
        return ;
    }

    int m=(l+r)>>1;
    if(pos<=m) update(pos,c,lson);
    if(pos> m) update(pos,c,rson);
    pushup(rt);
}

ll query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R){
         return tree[rt];
    }

    int m=(l+r)>>1;
    ll ret=inf;
    if(L<=m) ret=min(ret,query(L,R,lson));
    if(R> m) ret=min(ret,query(L,R,rson));
    return ret;
}

int main()
{
    int n,M,E;
    cin>>n>>M>>E;
    for(int i=1;i<=n;i++){
        cin>>a[i].l>>a[i].r>>a[i].val;
        a[i].l+=2,a[i].r+=2;
    }
    M+=2,E+=2;
    sort(a+1,a+1+n);
    build(1,maxn,1);
    update(M-1,0,1,maxn,1);
    for(int i=1;i<=n;i++){
        ll cnt=query(a[i].l-1,a[i].r,1,maxn,1);
        cnt+=a[i].val;
        update(a[i].r,cnt,1,maxn,1);
    }
    ll ans=query(E,a[n].r,1,maxn,1);
    if(ans<inf) cout<<ans<<endl;
    else cout<<"-1"<<endl;
}

OK了。