TOJ 2867.Picking Problem（最大区间调度）

题目链接：http://acm.tju.edu.cn/toj/showp2762.html

2867.   Picking Problem

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Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 1969   Accepted Runs: 831

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On holiday, Kandy happens to come to a square. People there are playing different kinds of games. Each game has a start time and a lasting duration. Two or more games may carry on at the same time. Kandy likes the games so much that he wants to take apart in as many as possible games. Given the list of all the games, can you help him to calculate the maximum number of games that he can take.

Input

There are several test cases. The first line of each test case is a single integer N, which is the total number of the games (1 ≤ N ≤ 10000). It's followed by N lines, each line has two integers, S and D, and S is the start time of the ith game, and D is the game's duration (0 ≤ S ≤ 10000, 1 ≤ D ≤ 1000).

The last case is followed by a single integer 0.

Output

For each case, output one line with a single integer M, which is the maximum number of games that Kandy can take.

Sample Input

50 35 109 23 42 20

Sample Output

3

Hint: In the sample, the first game starts at 0, ends at 3, and the forth game starts at 3, ends at 7, and the third game starts at 9, ends at 11. So he can pick these three games.

Source: TJU Team Selection Contest 2007 (3)

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题目大意：给定一系列活动的开始时间和结束时间，问最多能参加的活动数目

思路：// 本题属于最大区间调度问题，即数轴上有n个区间，选出最多的区间，使这些区间互相不重叠。算法：按右端点坐标排序，然后依次按后者的开始时间是否大于前者的结束时间（注意更新前者的下标）选择所有能选的区间。

#include <stdio.h>#include <algorithm>using namespace std;struct game{int s,e;};bool com(game a,game b){return a.e<b.e;}int main(){game games[10001];int n,dur;while(~scanf("%d",&n),n){for(int i=0;i<n;i++){scanf("%d%d",&games[i].s,&dur);games[i].e=games[i].s+dur;}sort(games,games+n,com);int m=0,sum=1;for(int i=1;i<n;i++)if(games[i].s>=games[m].e){sum++;m=i;}printf("%d\n",sum);}}