题目大意:
就是现在给出n <= 8000条线段, 都是垂直于x轴的且两两没有交点, 从中选出3条如果这3条两两能看到对方则算一种方案, 文友多少种方案
大致思路:
表示没有想到比较靠谱的计数方法, 用线段树处理出各个线段的可视情况之后如何计数感觉网上说的三重循环枚举的计数方法不可靠
但是还没有想到比较好的方法, 先水过去了...
注意纵坐标都乘上了2的原因是为了处理这一类情况:
前面有线段(2, 0)->(2, 2) 和线段(2, 3)->(2, 5)那么当出现(3, 1)->(3, 4)的时候这个线段应该要能通过那两个线段的缝隙看到前面的
代码如下:
Result : Accepted Memory : 63040 KB Time : 1500 ms
/* * Author: Gatevin * Created Time: 2015/8/17 17:57:22 * File Name: Sakura_Chiyo.cpp */ #include<iostream> #include<sstream> #include<fstream> #include<vector> #include<list> #include<deque> #include<queue> #include<stack> #include<map> #include<set> #include<bitset> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cstring> #include<cctype> #include<cmath> #include<ctime> #include<iomanip> using namespace std; const double eps(1e-8); typedef long long lint; #define maxn 8000*2 struct Segment_Tree { #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 int color[maxn << 2]; bool cansee[maxn/2][maxn/2]; void pushUp(int rt) { if(color[rt << 1] == color[rt << 1 | 1] && color[rt << 1] != -1) color[rt] = color[rt << 1]; else color[rt] = -1; } void pushDown(int rt) { if(color[rt] != -1) { color[rt << 1] = color[rt << 1 | 1] = color[rt]; color[rt] = -1; } } void update(int l, int r, int rt, int L, int R, int value) { if(l >= L && r <= R) { color[rt] = value; return; } int mid = (l + r) >> 1; pushDown(rt); if(mid >= L) update(lson, L, R, value); if(mid + 1 <= R) update(rson, L, R, value); pushUp(rt); } void query(int l, int r, int rt, int L, int R, int value) { if(l >= L && r <= R) { if(color[rt] != -1) { if(color[rt]) cansee[color[rt]][value] = cansee[value][color[rt]] = 1; return; } } int mid = (l + r) >> 1; pushDown(rt); if(mid >= L) query(lson, L, R, value); if(mid + 1 <= R) query(rson, L, R, value); pushUp(rt); } }; Segment_Tree ST; struct Segment { int x, y1, y2; Segment(int _x, int _y1, int _y2) { x = _x, y1 = _y1, y2 = _y2; } Segment(){} }; bool cmp(Segment s1, Segment s2) { return s1.x < s2.x; } Segment s[8080]; int main() { int T, n; scanf("%d", &T); while(T--) { scanf("%d", &n); int y1, y2, x; int ma = -1; for(int i = 1; i <= n; i++) { scanf("%d %d %d", &y1, &y2, &x); if(y1 > y2) swap(y1, y2); s[i] = Segment(x, y1*2, y2*2); ma = max(ma, y2*2); } sort(s + 1, s + n + 1, cmp); ST.color[1] = 0; memset(ST.cansee, 0, sizeof(ST.cansee)); for(int i = 1; i <= n; i++) { ST.query(0, ma, 1, s[i].y1, s[i].y2, i); ST.update(0, ma, 1, s[i].y1, s[i].y2, i); } int ans = 0; for(int i = 1; i <= n; i++) for(int j = i + 1; j <= n; j++) if(ST.cansee[i][j]) for(int k = j + 1; k <= n; k++) if(ST.cansee[i][k] && ST.cansee[j][k]) ans++; printf("%d\n", ans); } return 0; }