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：
1 /* 2 * author: yzcstc 3 * Created Time: 2013骞?9鏈?7鏃?鏄熸湡鍏?12鏃?7鍒?1绉? 4 * File Name: poj2481.cpp 5 */ 6 #include<iostream> 7 #include<sstream> 8 #include<vector> 9 #include<list> 10 #include<deque> 11 #include<queue> 12 #include<stack> 13 #include<map> 14 #include<set> 15 #include<algorithm> 16 #include<cstdio> 17 #include<cstdlib> 18 #include<cstring> 19 #include<cctype> 20 #include<cmath> 21 #define MXN 100010 22 #define MXM 1000000 23 #define M0(a) memset(a, 0, sizeof(a)) 24 #define eps 1e-8 25 #define Inf 0x7fffffff 26 using namespace std; 27 struct oo{ 28 int x, y, id; 29 bool operator <(const oo & b) const{ 30 return x < b.x || x == b.x && y > b.y; 31 } 32 }; 33 oo a[240000]; 34 int cnt[240000], n, ans[101010]; 35 36 int lowbit(int x){ 37 return x & (-x) ; 38 } 39 40 void updata(int x, int v){ 41 while (x <= MXN){ 42 cnt[x] += v; 43 x += lowbit(x); 44 } 45 } 46 47 int get_sum(int x){ 48 int ret = 0; 49 while (x){ 50 ret += cnt[x]; 51 x -= lowbit(x); 52 } 53 return ret; 54 } 55 56 bool cmp(const oo &a, const oo &b){ 57 if (a.y < b.y) return true; 58 if (a.y == b.y && a.x < b.x) return true; 59 return false; 60 } 61 62 void solve(){ 63 M0(cnt); 64 M0(ans); 65 for (int i = 1; i <= n; ++i){ 66 scanf("%d%d", &a[i].x, &a[i].y); 67 ++ a[i].x; 68 ++ a[i].y; 69 a[i].id = i; 70 } 71 sort(a + 1, a + n + 1); 72 int sum = 0; 73 for (int i = 1; i <= n; ++i){ 74 ans[a[i].id] = get_sum(a[i].y); 75 updata(a[i].x, 1); 76 updata(a[i].y, -1); 77 } 78 sort(a + 1, a + n + 1, cmp); 79 int l = 1, r = 1; 80 for (int i = 2; i <= n; ++i){ 81 while (l < i && a[l].y < a[i].y) ++l; 82 r = max(r, l); 83 while (r < i && a[r + 1].x < a[i].x) ++r; 84 if (a[r].x == a[i].x) r = i; 85 if (r < i) ans[a[i].id] += (r - l + 1); 86 } 87 for (int i = 1; i < n; ++i) 88 printf("%d ", ans[i]); 89 printf("%d ", ans[n]); 90 } 91 92 int main(){ 93 freopen("a.in","r",stdin); 94 freopen("a.out","w",stdout); 95 while (scanf("%d", &n) != EOF && n){ 96 solve(); 97 } 98 fclose(stdin); fclose(stdout); 99 return 0; 100 }
法2：
1 /* 2 * author: yzcstc 3 * Created Time: 2013骞?9鏈?7鏃?鏄熸湡鍏?12鏃?7鍒?1绉? 4 * File Name: poj2481.cpp 5 */ 6 #include<iostream> 7 #include<sstream> 8 #include<vector> 9 #include<list> 10 #include<deque> 11 #include<queue> 12 #include<stack> 13 #include<map> 14 #include<set> 15 #include<algorithm> 16 #include<cstdio> 17 #include<cstdlib> 18 #include<cstring> 19 #include<cctype> 20 #include<cmath> 21 #define MXN 100010 22 #define MXM 1000000 23 #define M0(a) memset(a, 0, sizeof(a)) 24 #define eps 1e-8 25 #define Inf 0x7fffffff 26 using namespace std; 27 struct oo{ 28 int x, y, id; 29 bool operator <(const oo & b) const{ 30 return x < b.x || x == b.x && y > b.y; 31 } 32 bool operator ==(const oo & b) const{ 33 return x == b.x && y == b.y; 34 } 35 }; 36 oo a[240000]; 37 int cnt[240000], n, ans[101010]; 38 39 int lowbit(int x){ 40 return x & (-x) ; 41 } 42 43 void updata(int x, int v){ 44 while (x <= MXN){ 45 cnt[x] += v; 46 x += lowbit(x); 47 } 48 } 49 50 int get_sum(int x){ 51 int ret = 0; 52 while (x){ 53 ret += cnt[x]; 54 x -= lowbit(x); 55 } 56 return ret; 57 } 58 59 void solve(){ 60 M0(cnt); 61 M0(ans); 62 for (int i = 1; i <= n; ++i){ 63 scanf("%d%d", &a[i].x, &a[i].y); 64 ++ a[i].x; 65 ++ a[i].y; 66 a[i].id = i; 67 } 68 sort(a + 1, a + n + 1); 69 for (int i = 1; i <= n; ++i){ 70 ans[a[i].id] = get_sum(MXN) - get_sum(a[i].y - 1); 71 updata(a[i].y, 1); 72 } 73 int l = 1; 74 for (int i = 2; i <= n; ++i){ 75 while (l < i){ 76 if (a[l] == a[i]) break; 77 ++l; 78 } 79 ans[a[i].id] -= (i - l); 80 } 81 for (int i = 1; i < n; ++i) 82 printf("%d ", ans[i]); 83 printf("%d ", ans[n]); 84 } 85 86 int main(){ 87 freopen("a.in","r",stdin); 88 freopen("a.out","w",stdout); 89 while (scanf("%d", &n) != EOF && n){ 90 solve(); 91 } 92 fclose(stdin); fclose(stdout); 93 return 0; 94 }