POJ 1436 Horizontally Visible Segments (线段树+区间覆盖)

题意：给定一个二维平面，再给你n条垂直线条，如果有两条线段能用水平的线连接且不水平的线不接触其他线段，则说明这条两条线段可见，如果三条线段两两可见，则代表一个“三角形”，问给定线段集合中的“三角形”个数，

解题思路：线段树区间覆盖加统计，要把区间变为点数，所以所有点的 y 都要 乘 2 。 不会stl的童鞋伤不起，用链表统计，更新，最后在暴力找个数。

解题代码：

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <stdlib.h>
  4 #define  maxn 8005
  5 int  hs[maxn];
  6 struct op
  7 {
  8     int y1,y2,x;
  9 } ops[maxn];
 10 int cmp(const void *a , const void *b)
 11 {
 12     return (*(op*)a).x - (*(op*)b).x;
 13 }
 14 struct li
 15 {
 16     int num ;
 17     struct li*next;
 18 }ths[maxn];
 19 struct li * newli()
 20 {
 21     struct li *p = (li*)malloc(sizeof(li));
 22     p->num = 0 ;
 23     p->next = NULL;
 24     return p ;
 25 }
 26 struct node
 27 {
 28     int l , r, m ;
 29     int c ;
 30     struct li head;
 31     struct li* last;
 32 } tree[maxn*8];
 33 int L(int c)
 34 {
 35     return 2*c;
 36 }
 37 int R(int c)
 38 {
 39     return 2*c + 1;
 40 }
 41 void Free(struct li *p)
 42 {
 43     if(p->next != NULL)
 44        Free(p->next);
 45      free(p);
 46 }
 47 void build(int c, int p , int v)
 48 {
 49     tree[c].l = p ;
 50     tree[c].r = v;
 51     tree[c].m = (p + v)/2;
 52     tree[c].c = 0 ;
 53     tree[c].head.next = NULL;
 54     tree[c].head.num = 0 ;
 55     tree[c].last = &tree[c].head;
 56     if(p == v)
 57         return ;
 58     build(L(c),p,tree[c].m);
 59     build(R(c),tree[c].m+1,v);
 60 }
 61 
 62 void Pushdown(int c)
 63 {
 64 
 65     if(tree[c].c != 0)
 66     {
 67        tree[L(c)].c = tree[c].c;
 68         tree[R(c)].c = tree[c].c;
 69         struct li *p;
 70         p = newli();
 71         p->num = tree[c].c;
 72         tree[L(c)].head.num = 1;
 73         tree[L(c)].head.next = p ;
 74         tree[L(c)].last = p;
 75         p = newli();
 76         p->num = tree[c].c;
 77         tree[R(c)].head.num = 1;
 78         tree[R(c)].head.next = p ;
 79         tree[R(c)].last = p;
 80         tree[c].c  = 0;
 81     }
 82 }
 83 
 84 void Pushup(int c)
 85 {
 86     tree[c].head.num = tree[L(c)].head.num + tree[R(c)].head.num;
 87     tree[L(c)].last->next = tree[R(c)].head.next;
 88     tree[c].head.next = tree[L(c)].head.next;
 89     if(tree[c].head.num == 0 )
 90       tree[c].last = &tree[c].head;
 91     else if(tree[R(c)].head.num == 0)
 92       tree[c].last = tree[L(c)].last;
 93     else tree[c].last = tree[R(c)].last;
 94 }
 95 void update(int c, int p , int v, int value)
 96 {
 97     if(p <= tree[c].l &&v >= tree[c].r)
 98     {
 99         tree[c].c = value;
100         struct li *p = &tree[c].head;
101         for(int i = 1; i <= tree[c].head.num; i ++)
102         {
103             p = p->next ;
104             hs[p->num] = 1;
105         }
106         p = newli();
107         p->num = value;
108         tree[c].head.num = 1;
109         tree[c].head.next = p ;
110         tree[c].last = p;
111         return ;
112     }
113     Pushdown(c);
114     if(v <= tree[c].m) update(L(c),p,v,value);
115     else if(p > tree[c].m) update(R(c),p,v,value);
116     else
117     {
118         update(L(c),p,tree[c].m,value);
119         update(R(c),tree[c].m+1,v,value);
120     }
121     Pushup(c);
122 }
123 int sum =0 ,n;
124 void fuck(int k)
125 {
126   for(int i = n;i >= 1;i --)
127   {
128       if(hs[i] == 1)
129       {
130         ths[k].num++;
131         struct li *p = newli();
132         p->num = i ;
133         p->next = ths[k].next ;
134         ths[k].next = p ;
135       }
136   }
137 }
138 void F(int k)
139 {
140     struct li * p1 ,*p2 ;
141     p1 = &ths[k];
142     bool visit[maxn] = {0};
143     for(int i = 1; i <= ths[k].num ; i++)
144     {
145             p1 = p1->next;
146             visit[p1->num] = 1;
147             p2 = &ths[p1->num];
148             for(int j= 1; j <= ths[p1->num].num; j ++)
149             {
150                   p2 = p2->next;
151                   if(visit[p2->num])
152                     sum ++;
153             }
154     }
155 }
156 int main()
157 {
158     int t ;
159     scanf("%d",&t);
160     while(t--)
161     {
162         sum = 0 ;
163         memset(tree,0,sizeof(tree));
164         memset(ths,0,sizeof(ths));
165         scanf("%d",&n);
166         for(int i =1 ; i <= n; i ++)
167             scanf("%d %d %d",&ops[i].y1,&ops[i].y2,&ops[i].x);
168         qsort(ops +1 , n, sizeof(op),cmp);
169         build(1,0,2*maxn);
170         for(int i= 1; i <= n; i ++)
171         {
172             memset(hs,0,sizeof(hs));
173             update(1,ops[i].y1*2,ops[i].y2*2,i);
174             fuck(i);
175             F(i);
176         }
177         printf("%d\n",sum);
178     }
179     return 0;
180 }

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