Cleaning Shifts
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4245 | Accepted: 1429 |
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
题意：给n个区间及其代价值，问要覆盖[M,E]区间至少要花费多少代价；
解法：这是一个dp问题，先列出方程。
F[i]表示取[0,i]这个区间的代价,初始化F[M-1]=0,答案就是F[E].
则方程为F[a[i].T2]=min(F[a[j].T2])+a[i].s (T1-1<=a[j].T2<T2),找min的过程用线段树实现。
将a[i]按T2从小到大排列,逐步更新最小值。
代码：
1 #include"bits/stdc++.h"
2
3 #define ll long long
4 #define vl vector<ll>
5 #define ci(x) scanf("%d",&x)
6 #define pi(x) printf("%d\n",x)
7 #define pl(x) printf("%lld\n",x)
8 #define rep(i, n) for(int i=0;i<n;i++)
9 using namespace std;
10 const int NN = 1e6 + 5;
11 int n,s,t;
12 struct P{int x,y,s;};
13 P a[NN];
14 bool cmp(P a,P b){
15 return a.y<b.y;
16 }
17 const ll INF = 0x3fffffffffffffff;
18 struct SegMin {
19 int N;
20 vl is;vl mul;vl add;
21 ll init;
22 ll merge(ll a, ll b) {
23 return min(a, b);
24 }
25 void push(int o, int L, int R, ll m, ll a) {
26 is[o] = is[o] * m + a;
27 mul[o] = mul[o] * m;
28 add[o] = add[o] * m + a;
29 }
30
31 SegMin(int n, ll init=INF) {
32 N = 1;
33 while (N < n) N *= 2;
34 this->init = init;
35 is = vl(N * 4, init);
36 mul = vl(N * 4, 1);
37 add = vl(N * 4);
38 }
39
40 SegMin(vl a, ll init=INF) {
41 int n = a.size();
42 N = 1;
43 while (N < n) N *= 2;
44 this->init = init;
45 is = vl(N * 2);
46 mul = vl(N * 2, 1);
47 add = vl(N * 2);
48 copy(a.begin(), a.end(), is.begin() + N);
49 for (int i = N - 1; i > 0; i--) {
50 is[i] = merge(is[i << 1], is[i << 1 | 1]);
51 }
52 }
53
54 void update(int l, int r, ll m, ll a) {
55 if (l < r) update(1, 0, N, l, r, m, a);
56 }
57
58 void update(int o, int L, int R, int l, int r, ll m, ll a) {
59 if (l <= L && R <= r) {
60 push(o, L, R, m, a);
61 } else {
62 int M = (L + R) >> 1;
63 push(o, L, M, R);
64 if (l < M) update(o << 1, L, M, l, r, m, a);
65 if (r > M) update(o << 1 | 1, M, R, l, r, m, a);
66 is[o] = merge(is[o << 1], is[o << 1 | 1]);
67 }
68 }
69
70 void push(int o, int L, int M, int R) {
71 if (mul[o] != 1 || add[o] != 0) {
72 push(o << 1, L, M, mul[o], add[o]);
73 push(o << 1 | 1, M, R, mul[o], add[o]);
74 mul[o] = 1;
75 add[o] = 0;
76 }
77 }
78
79 ll query(int l, int r) {
80 if (l < r) return query(1, 0, N, l, r);
81 return init;
82 }
83
84 ll query(int o, int L, int R, int l, int r) {
85 if (l <= L && R <= r) {
86 return is[o];
87 } else {
88 int M = (L + R) >> 1;
89 push(o, L, M, R);
90 ll res = init;
91 if (l < M) res = merge(res, query(o << 1, L, M, l, r));
92 if (r > M) res = merge(res, query(o << 1 | 1, M, R, l, r));
93 is[o] = merge(is[o << 1], is[o << 1 | 1]);
94 return res;
95 }
96 }
97 };
98
99 int main(){
100 ci(n),ci(s),ci(t);//s从1开始
101 s++,t++;
102 int ma=0;
103 for(int i=0;i<n;i++) ci(a[i].x),ci(a[i].y),ci(a[i].s);
104 for(int i=0;i<n;i++) a[i].x++,a[i].y++,ma=max(ma,a[i].y);
105 sort(a,a+n,cmp);
106 SegMin seg(ma+1);
107 seg.update(0,ma+1,0,INF);
108 seg.update(0,s,0,0);
109
110 for(int i=0;i<n;i++){
111 if(a[i].y<s) continue;
112 int L=a[i].x-1,R=a[i].y;
113 ll res=seg.query(L,R)+a[i].s;
114 res=min(seg.query(R,R+1),res);//与前面的最小值取min
115 seg.update(R,R+1,0,res);
116 }
117 ll ans=seg.query(t,ma+1);
118 if(ans>=INF) puts("-1");//未覆盖到
119 else pl(ans);
120 return 0;
121 }
原文地址：https://www.cnblogs.com/mj-liylho/p/9502012.html