poj2481

题意：给定一些线段(s, e),起点为s，终点为e，求每一段线段被多少线段包含（不包括相等）

思路：很明显的树状数组题目。。但是做的时候想了挺久。。（下面的x为线段起点， y为线段终点）

做法1：先对线段进行排序，比较函数为 a.x < b.x || a.x == b.x && a.y > b.y;

接下来便依次插入树状数组中，插的时候左端点 +1， 右端点-1，这样求和时前面的线段自然消掉

统计是算sum(a[i].y)即可。。

但是这样我们发现落下了一种情况，就是把 x < a[i].x && y == a[i].y情况给消掉了，所以还要排序在统计一遍。。

这种做法我写完2300+ms过了。。

做法2：差不多的思路，但是结合把(x, y)当成二维坐标，这样我们发现其实上就是求其左上方的点有多少个。。采用poj2352stars做法即可。。需要判重。。

以为会快很多，。。没想到也要2200+ms才过。。

法1：

 /*
  * author:  yzcstc
  * Created Time:  2013骞?9鏈?7鏃?鏄熸湡鍏?12鏃?7鍒?1绉?
  * File Name: poj2481.cpp
  */
 #include<iostream>
 #include<sstream>
 #include<vector>
 #include<list>
 #include<deque>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<algorithm>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cctype>
 #include<cmath>
 #define MXN 100010
 #define MXM 1000000
 #define M0(a) memset(a, 0, sizeof(a))
 #define eps 1e-8
 #define Inf 0x7fffffff
 using namespace std;
 struct oo{
     int x, y, id;
     bool operator <(const oo & b) const{
           return x < b.x || x == b.x && y > b.y;
     }
 };
 oo a[];
 int cnt[], n, ans[];

 int lowbit(int x){
     return x & (-x) ;
 }

 void updata(int x, int v){
     while (x <= MXN){
          cnt[x] += v;
          x += lowbit(x);
     }
 }

 int get_sum(int x){
     int ret = ;
     while (x){
         ret += cnt[x];
         x -= lowbit(x);
     }
     return ret;
 }

 bool cmp(const oo &a, const oo &b){
      if (a.y < b.y) return true;
      if (a.y == b.y && a.x < b.x) return true;
      return false;
 }

 void solve(){
     M0(cnt);
     M0(ans);
     for (int i = ; i <= n; ++i){
         scanf("%d%d", &a[i].x, &a[i].y);
         ++ a[i].x;
         ++ a[i].y;
         a[i].id = i;
     }
     sort(a + , a + n + );
     int sum = ;
     for (int i = ; i <= n; ++i){
        ans[a[i].id] = get_sum(a[i].y);
        updata(a[i].x, );
        updata(a[i].y, -);
     }
     sort(a + , a + n + , cmp);
     int l = , r = ;
     for (int i = ; i <= n; ++i){
         while (l < i && a[l].y < a[i].y) ++l;
         r = max(r, l);
         while (r < i && a[r + ].x < a[i].x) ++r;
         if (a[r].x == a[i].x) r = i;
         if (r < i) ans[a[i].id] +=  (r - l + );
     }
     for (int i = ; i < n; ++i)
          printf("%d ", ans[i]);
     printf("%d\n", ans[n]);
 }

 int main(){
     freopen("a.in","r",stdin);
     freopen("a.out","w",stdout);
     while (scanf("%d", &n) != EOF && n){
         solve();
     }
     fclose(stdin);  fclose(stdout);
     return ;
 }

法2：

 /*
  * author:  yzcstc
  * Created Time:  2013骞?9鏈?7鏃?鏄熸湡鍏?12鏃?7鍒?1绉?
  * File Name: poj2481.cpp
  */
 #include<iostream>
 #include<sstream>
 #include<vector>
 #include<list>
 #include<deque>
 #include<queue>
 #include<stack>
 #include<map>
 #include<set>
 #include<algorithm>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cctype>
 #include<cmath>
 #define MXN 100010
 #define MXM 1000000
 #define M0(a) memset(a, 0, sizeof(a))
 #define eps 1e-8
 #define Inf 0x7fffffff
 using namespace std;
 struct oo{
     int x, y, id;
     bool operator <(const oo & b) const{
           return x < b.x || x == b.x && y > b.y;
     }
     bool operator ==(const oo & b) const{
           return x == b.x && y == b.y;
     }
 };
 oo a[];
 int cnt[], n, ans[];

 int lowbit(int x){
     return x & (-x) ;
 }

 void updata(int x, int v){
     while (x <= MXN){
          cnt[x] += v;
          x += lowbit(x);
     }
 }

 int get_sum(int x){
     int ret = ;
     while (x){
         ret += cnt[x];
         x -= lowbit(x);
     }
     return ret;
 }

 void solve(){
     M0(cnt);
     M0(ans);
     for (int i = ; i <= n; ++i){
         scanf("%d%d", &a[i].x, &a[i].y);
         ++ a[i].x;
         ++ a[i].y;
         a[i].id = i;
     }
     sort(a + , a + n + );
     for (int i = ; i <= n; ++i){
        ans[a[i].id] = get_sum(MXN) - get_sum(a[i].y - );
        updata(a[i].y, );
     }
     int l = ;
     for (int i = ; i <= n; ++i){
        while (l < i){
              if (a[l] == a[i])  break;
              ++l;
        }
        ans[a[i].id] -= (i - l);
     }
     for (int i = ; i < n; ++i)
          printf("%d ", ans[i]);
     printf("%d\n", ans[n]);
 }

 int main(){
     freopen("a.in","r",stdin);
     freopen("a.out","w",stdout);
     while (scanf("%d", &n) != EOF && n){
         solve();
     }
     fclose(stdin);  fclose(stdout);
     return ;
 }