1420 - Priest John's Busiest Day （贪心）
John is the only priest in his town. October 26th is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i -th couple plan to hold their wedding from time Si to time Ti . According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. Moreover, this ceremony must be longer than half of the wedding time and can't be interrupted. Could you tell John how to arrange his schedule so that he can hold all special ceremonies of all weddings?
Please note that:
John can not hold two ceremonies at the same time.
John can only join or leave the weddings at integral time.
John can show up at another ceremony immediately after he finishes the previous one.
Input
The input consists of several test cases and ends with a line containing a zero. In each test case, the first line contains a integer N (1N100, 000) indicating the total number of the weddings. In the next Nlines, each line contains two integers Si and Ti ( 0Si < Ti2147483647 ).
Output
For each test, if John can hold all special ceremonies, print ``YES"; otherwise, print ``NO".
Sample Input
3 1 5 2 4 3 6 2 1 5 4 6 0
Sample Output
NO YES
题意：有n场婚礼，开始时间s结束时间e，每场婚礼要花超过一半时间做仪式，问能否所有婚礼都成功仪式。
哎。这题一开是想残了。一直想着用贪心+优先队列去做，wa了。。后面仔细想想才发现想法是错的。
思路：按每场婚礼最晚可以仪式的时间mid排序。然后维护一个Min作为区间左，然后每次判断是否超过mid。这样做的可行性建立在，每个区间即使从s开始，到他仪式完的时间肯定超过mid。所以mid越小的肯定要越先去完成。
代码：
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define max(a,b) (a)>(b)?(a):(b)
using namespace std;
const int N = 100005;
int n;
struct D {
int s, e, mid;
} d[N];
bool cmp(D a, D b) {
return a.mid < b.mid;
}
void init() {
for (int i = 0; i < n; i++) {
scanf("%d%d", &d[i].s, &d[i].e);
d[i].mid = d[i].e - ((d[i].e - d[i].s) / 2 + 1);
}
}
bool judge() {
int Min = 0;
for (int i = 0; i < n; i++) {
if (d[i].mid < Min) return false;
Min = max(Min, d[i].s);
Min += (d[i].e - d[i].s) / 2 + 1;
}
return true;
}
void solve() {
sort(d, d + n, cmp);
if (judge()) printf("YES\n");
else printf("NO\n");
}
int main() {
while (~scanf("%d", &n) && n) {
init();
solve();
}
return 0;
}