POJ1769_Minimizing maximizer_DP|線段樹最佳化

Minimizing maximizer

Time Limit: 5000MS		Memory Limit: 30000K
Total Submissions: 4355		Accepted: 1792

Description The company Chris Ltd. is preparing a new sorting hardware called Maximizer. Maximizer has n inputs numbered from 1 to n. Each input represents one integer. Maximizer has one output which represents the maximum value present on Maximizer's inputs. 

Maximizer is implemented as a pipeline of sorters Sorter(i1, j1), ... , Sorter(ik, jk). Each sorter has n inputs and n outputs. Sorter(i, j) sorts values on inputs i, i+1,... , j in non-decreasing order and lets the other inputs pass through unchanged. The n-th output of the last sorter is the output of the Maximizer. 

An intern (a former ACM contestant) observed that some sorters could be excluded from the pipeline and Maximizer would still produce the correct result. What is the length of the shortest subsequence of the given sequence of sorters in the pipeline still producing correct results for all possible combinations of input values? 

Task 
Write a program that: 

reads a description of a Maximizer, i.e. the initial sequence of sorters in the pipeline, 
computes the length of the shortest subsequence of the initial sequence of sorters still producing correct results for all possible input data, 
writes the result. 

Input The first line of the input contains two integers n and m (2 <= n <= 50000, 1 <= m <= 500000) separated by a single space. Integer n is the number of inputs and integer m is the number of sorters in the pipeline. The initial sequence of sorters is described in the next m lines. The k-th of these lines contains the parameters of the k-th sorter: two integers ik and jk (1 <= ik < jk <= n) separated by a single space.

Output The output consists of only one line containing an integer equal to the length of the shortest subsequence of the initial sequence of sorters still producing correct results for all possible data.

Sample Input

40 620 301 1010 2020 3015 2530 40

Sample Output

4

Hint Huge input data, scanf is recommended.

#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int maxn = 50000 + 1000;const int inf  = 0x3f3f3f3f;int Seg[maxn * 4];//線段樹維護區間最小值int n, m;//線段樹查詢區間最小值int Query(int p, int l, int r, int x, int y){if(l>= x && r <= y) return Seg[p];int mid = (l + r) >> 1;int res = inf;if(x <= mid) res = Query(p<<1, l, mid, x, y);if(y > mid) res = min(res, Query(p<<1|1, mid+1, r, x, y));return res;}//線段樹單點更新void Update(int p, int l, int r, int x, int y){if(l == r){Seg[p] = min(y, Seg[p]);return;}int mid = (l + r) >> 1;if(x <= mid) Update(p<<1, l, mid, x, y);else Update(p<<1|1, mid+1, r, x, y);Seg[p] = min(Seg[p<<1], Seg[p<<1|1]);}int main(){//線段樹初始化為無窮大memset(Seg, inf, sizeof(Seg));scanf("%d %d", &n, &m);// dp[1] 初始化為 0Update(1, 1, n, 1, 0);while(m--){int s, t;scanf("%d %d", &s, &t);//狀態轉移int temp = Query(1, 1, n, s, t) + 1;Update(1, 1, n, t, temp);}printf("%d\n", Query(1, 1, n, n, n));return 0;}