思维+multiset ZOJ Monthly, July 2015 - H Twelves Monkeys

题目传送门

 /*
     题意：n个时刻点，m次时光穿梭，告诉的起点和终点，q次询问，每次询问t时刻t之前有多少时刻点是可以通过两种不同的路径到达
     思维：对于当前p时间，从现在到未来穿越到过去的是有效的值，排个序，从大到小询问，那么之前添加的穿越点都是有效的，
         用multiset保存。比赛时想到了排序，但是无法用线段树实现查询，stl大法好！
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <set>
 #include <iostream>
 using namespace std;
 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 struct Year {
     int u, v;
     bool operator < (const Year &r) const   {
         return u > r.u;
     }
 }y[MAXN];
 struct Query   {
     int p, id;
     bool operator < (const Query &r) const  {
         return p > r.p;
     }
 }q[MAXN];
 int ans[MAXN];
 int a[];
 int n, m, l;
 int main(void)  {       //ZOJ Monthly, July 2015 - H Twelves Monkeys
     //freopen ("H.in", "r", stdin);
     while (scanf ("%d%d%d", &n, &m, &l) == )   {
         for (int i=; i<=m; ++i)    {
             scanf ("%d%d", &y[i].u, &y[i].v);
         }
         for (int i=; i<=l; ++i)    {
             scanf ("%d", &q[i].p);  q[i].id = i;
         }
         sort (y+, y++m);  sort (q+, q++l);
         multiset<int> S;    int j = ;
         for (int i=; i<=l; ++i)    {
             while (j <= m && y[j].u >= q[i].p)  {
                 S.insert (y[j].v);  j++;
             }
             multiset<int>::iterator it; int cnt = ;
             for (it=S.begin (); it!=S.end (); ++it) {
                 a[++cnt] = *(it);   if (cnt == )   break;
             }
             if (cnt ==  && a[] <= q[i].p) ans[q[i].id] = q[i].p - a[];
             else    ans[q[i].id] = ;
         }
         for (int i=; i<=l; ++i)
             printf ("%d\n", ans[i]);
     }
     return ;
 }