线段树入门---给定多个线段求点的出现个数

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <algorithm>
  4 
  5 using namespace std;
  6 
  7 const int MAX = 999999999;
  8 const int MIN = 0;
  9 //left，right表示区间的左右端点，cnt表示这样的线段有多少条
 10 typedef struct Node{
 11     int left, right;
 12     int cnt;
 13     struct Node *lchild, *rchild;
 14 }Node, *PNode;
 15 //创建线段树函数
 16 void CreateTree(PNode *root)//*root已经被初始化
 17 {
 18     int left = (*root)->left;
 19     int right = (*root)->right;
 20     if(left < right)
 21     {
 22         int mid = (left + right) / 2;
 23         //它的左孩子,并初始化它的左孩子
 24         PNode lnode = (PNode)malloc(sizeof(Node));
 25         lnode->cnt = 0;
 26         lnode->left = left;
 27         lnode->right = mid;
 28         lnode->lchild = lnode->rchild = NULL;
 29         //它的右孩子,并初始化它的右孩子
 30         PNode rnode = (PNode)malloc(sizeof(Node));
 31         rnode->cnt = 0;
 32         rnode->left = mid + 1;
 33         rnode->right = right;
 34         rnode->rchild = rnode->lchild = NULL;
 35         //扯上关系
 36         (*root)->lchild = lnode;
 37         (*root)->rchild = rnode;
 38         //递归建树
 39         CreateTree(&(*root)->lchild);
 40         CreateTree(&(*root)->rchild);
 41     }
 42 }
 43 //将区间映射到线段树上
 44 void Map_Interval(PNode root, int left, int right)
 45 {
 46     //如果恰好当前区间等于线段树当前节点的区间，将该条线段+1，就返回
 47     if(root->left == left && root->right == right)
 48     {
 49         root->cnt++;
 50         return;
 51     }
 52     int mid = (root->left + root->right) / 2;
 53     //如果在当前节点的区间左半个区间内， 就继续找他的左半个区间，也就是左儿子
 54     if(right <= mid)
 55     {
 56         Map_Interval(root->lchild, left, right);
 57     }
 58     //如果在当前节点的区间右半个区间内， 就继续找他的右半个区间，也就是右儿子
 59     else if(left > mid)
 60     {
 61         Map_Interval(root->rchild, left, right);
 62     }
 63     //如果左端点在当前节点的左半部份，右端点在当前区间的右半部份，分别找
 64     else
 65     {
 66         Map_Interval(root->lchild, left, mid);
 67         Map_Interval(root->rchild, mid + 1, right);
 68     }
 69 }
 70 //找到这个点在线段上一共出现的次数， sum用来得到具体的数目，  point为点的号
 71 void search_point(PNode root, int point, int *sum)
 72 {
 73     if(point == root->left && point == root->right)
 74     {
 75         *sum += root->cnt;
 76         return;
 77     }
 78     else
 79     {
 80         int mid = (root->left + root->right) / 2;
 81         if(point > mid)
 82         {
 83             *sum += root->cnt;
 84             search_point(root->rchild, point, sum);
 85         }
 86         else
 87         {
 88             *sum += root->cnt;
 89             search_point(root->lchild, point, sum);
 90         }
 91     }
 92 }
 93 //前序遍历，辅助看建树的情况
 94 void preOrder(PNode root)
 95 {
 96     if(root != NULL)
 97     {
 98         printf("[%d, %d]   count --> %d\n", root->left, root->right, root->cnt);
 99         preOrder(root->lchild);
100         preOrder(root->rchild);
101     }
102 }
103 
104 int main()
105 {
106     freopen("1.in", "r", stdin);
107     //初始化root
108     PNode root = (PNode)malloc(sizeof(Node));
109     root->cnt = 0;
110     root->lchild = root->rchild = NULL;
111 
112     int left[1000], right[1000], point[1000];
113     int n; //区间个数
114     scanf("%d", &n);
115     int min_num = MAX;//保存最小的区间端点
116     int max_num = MIN;//保存最大的区间端点
117     for(int i = 0; i < n; i++)
118     {
119         scanf("%d %d", &left[i], &right[i]);
120         min_num = min(left[i], min_num);
121         max_num = max(right[i], max_num);
122     }
123     int m;//带查询点的个数
124     scanf("%d", &m);
125     for(int i = 0; i < m; i++)
126         scanf("%d", &point[i]);
127     root->left = min_num;
128     root->right = max_num;
129     CreateTree(&root);//建树
130     preOrder(root);//打印出来，看建树的情况
131     for(int i = 0; i < n; i++)
132     {
133         Map_Interval(root, left[i], right[i]);
134     }
135     //将区间映射到树上之后，再看树的情况
136     printf("\n=================================================\n");
137     preOrder(root);
138     int tmp = 0;
139     for(int i = 0; i < m; i++)
140     {
141         search_point(root, point[i], &tmp);
142         printf("%d 在给定线段中出现 %d 次\n", point[i], tmp);
143         tmp = 0;
144     }
145     return 0;
146 }