UVa 1420 - Priest John's Busiest Day (贪心)
题意
有N对人要举行婚礼,从S到T,牧师必须参加一半以上 的婚礼过程,问他能不能参加完全部的婚礼。
思路
可以想到如果牧师赶到某个婚礼的时候的时间超过了MID值,他就完不成了。所以只要求出所有婚礼的MID值,排序,然后贪心即可。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-6;
const int MAXN = 1e5 + 10;
const int MOD = 1000007;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct POINT
{
int st, ed, mid;
bool operator < (const POINT &a) const
{
if (mid != a.mid) return mid < a.mid;
return st < a.st;
}
}pit[MAXN];
int main()
{
//ROP;
int n, i, j;
while (scanf("%d", &n), n)
{
for (i = 0; i < n; i++)
{
scanf("%d%d", &pit[i].st, &pit[i].ed);
pit[i].mid = pit[i].st + (pit[i].ed - pit[i].st) / 2 + 1;
}
sort(pit, pit + n);
int curTime = 0;
for (i = 0; i < n; i++)
{
if (curTime >= pit[i].mid) break;
curTime = max(curTime, pit[i].st);
curTime += pit[i].mid - pit[i].st;
}
printf("%s\n", i == n ? "YES" : "NO");
}
return 0;
}