poj3171 Cleaning Shifts【线段树（单点修改区间查询）】【DP】

Cleaning Shifts

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4422		Accepted: 1482

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：

一个[L, R]的区间，有n头牛，每头牛可以清理一个固定区间，需要花费一定的价格。现在要清理[L,R]这个区间，需要花费最少的价格是多少。

思路：

用f[x]表示清理区间[L, x]需要花费的最小价格。

对于一头牛，他可以清理的区间是[a,b]，那么因为中间不能间断，所以f[b] = min(f[x]) + c其中x是属于[a-1,b]区间的。

每次状态转移都要取一个区间最小值，并且更新一个点的值，所以用上线段树来维护。

首先将所有的牛按照结束的区间进行排序，然后DP。

因为时间是从0开始，所以我给所有的时间都加了2，这样a-1就还是从1开始。

要注意的是update（）时，并不是直接将值改为val，而是取较小值，这里WA了一会。

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f

 const int maxn = 1e5 + ;
 const int maxtime = ;
 struct node{
     int st, ed, cost;
 }cow[maxn];
 bool cmp(node a, node b)
 {
     return a.ed < b.ed;
 }
 LL tree[maxn << ];//区间中f[]最小值
 int n, L, R;

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

 void build(int rt, int l, int r)
 {
     if(l == r){
         tree[maxn] = inf;
         return;
     }
     int mid = (l + r) / ;
     build(rt<<, l, mid);
     build(rt<<|, mid + , r);
     pushup(rt);
 }

 void update(int x, LL val, int l, int r, int rt)
 {
     if(l == r){
         tree[rt] = min(tree[rt], val);
         return;
     }
     int m = (l + r) / ;
     if(x <= m){
         update(x, val, l, m, rt<<);
     }
     else{
         update(x, val, m + , r, rt<<|);
     }
     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) / ;
     LL ans = inf;
     if(L <= m){
         ans = min(ans, query(L, R, l, m, rt<< ));
     }
     if(R > m){
         ans = min(ans, query(L, R, m + , r, rt<<|));
     }
     pushup(rt);
     return ans;
 }

 int main()
 {
     while(scanf("%d%d%d", &n, &L, &R) != EOF){
         L+=;R+=;
         memset(tree, 0x7f, sizeof(tree));
         for(int i = ; i <= n; i++){
             scanf("%d%d%d", &cow[i].st, &cow[i].ed, &cow[i].cost);
             cow[i].st+=;cow[i].ed+=;
         }
         sort(cow + , cow +  + n, cmp);

         build(, , R);

         update(L - , , , R, );
         //cout<<"yes"<<endl;
         int far = L;
         bool flag = true;
         for(int i = ; i <= n; i++){
             if(cow[i].st > far + ){
                 flag = false;
             //    break;
             }
             int a = max(L - , cow[i].st - );
             int b = min(R, cow[i].ed);
             //cout<<a<<" "<<b<<endl;
             LL f = query(a, b, , R, );
             f += cow[i].cost;
             //cout<<f<<endl;
             update(b, f, , R, );
             far = max(far, cow[i].ed);
             //cout<<far<<endl;
         }
         //cout<<"yes"<<endl;

         LL ans = query(R, R, , R, );
         if(ans >= inf){
             printf("-1\n");
         }
         else{
                 printf("%lld\n", ans);

         //else{
         //    printf("-1\n");
         }

     }

 }