UVaLive 6609 Meeting Room Arrangement (贪心，区间不相交)

题意：给定 n 个区间，让你选出最多的区间，使得每个区间不相交。

析：贪心题，贪心策略是按右端点排序，然后按着选即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {1, 0, -1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
struct node{
    int s, f;
    bool operator < (const node &p) const{
        return f < p.f || f == p.f && s < p.s;
    }
};
node a[25];

int main(){
    int T;  cin >> T;
    while(T--){
        int x, y;
        int cnt = 0;
        while(true){
            scanf("%d %d", &x, &y);
            if(!x && !y) break;
            a[cnt].s = x;
            a[cnt++].f = y;
        }
        sort(a, a+cnt);
        int ans = 0, e = -1;
        for(int i = 0; i < cnt; ++i){
            if(a[i].s >= e){
                ++ans;
                e = a[i].f;
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}