HDU 3627【线段树变形+离散化】

题目：Giant For

题意：

一个容器有三种操作：

add x y: 增加一坐标。

remove x y: 去除一坐标。

find x y: 查找容器里的一坐标使得其座标值大于x y，如果有多组值，按主以x次以y排序，输出最小x, y

(0 <= x, y<= 1000000000)

解题思路:

用线段树维护坐标点，以x值作为叶结点，然后第个结点保存其子树的最大y值，如果find x, y,从根结点开始，往符合条件的子树深入，即往子树有大于x的叶结点和大于y值的结点的子树前进

这里x,y值很大，数组开不下去，但操作数最多才200000，也就是最多才有200000对不同的数，首选输入所有的操作，按x从小到大排序数值，可以以坐标号代替x，这就是离散化了，很多时候都很有用。处理的时候要注意相同值的影响。

View Code

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <string>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <map>
  8 
  9 using namespace std;
 10 
 11 const int MAX = 200000 + 10;
 12 struct TTree
 13 {
 14     int left, right, maxCol;
 15 }Tree[4 * MAX];
 16 
 17 struct TCoord
 18 {
 19     bool operator == (const TCoord& b)
 20     {
 21         return row == b.row && col == b.col;
 22     }
 23     int row, col;
 24 }Coord[MAX];
 25 
 26 int Hash[MAX];
 27 int Pos[MAX];
 28 char OP[MAX][10];
 29 
 30 bool comp(int i, int j)
 31 {
 32     if(Coord[i].row != Coord[j].row)
 33         return Coord[i].row < Coord[j].row;
 34     else
 35         return Coord[i].col < Coord[j].col;
 36 }
 37 
 38 int myMax(int a, int b)
 39 {
 40     return a < b ? b : a;
 41 }
 42 
 43 void build(int left, int right, int num)
 44 {
 45     Tree[num].maxCol = 0;
 46     Tree[num].left = left;
 47     Tree[num].right = right;
 48     if(left == right)
 49         return;
 50     int mid = (left + right) >> 1;
 51     build(left, mid, 2 * num);
 52     build(mid + 1, right, 2 * num + 1);
 53 }
 54 
 55 void insert(int row, int col, int num)
 56 {
 57     if(Tree[num].left == Tree[num].right)
 58     {
 59         Tree[num].maxCol = col;
 60         return;
 61     }
 62     int mid = (Tree[num].left + Tree[num].right) >> 1;
 63     if(row <= mid)
 64         insert(row, col, 2 * num);
 65     else
 66         insert(row, col, 2 * num + 1);
 67     Tree[num].maxCol = myMax(Tree[2 * num].maxCol, Tree[2 * num + 1].maxCol);
 68 }
 69 
 70 void remove(int row, int col, int num)
 71 {
 72     if(Tree[num].left == Tree[num].right)
 73     {
 74         Tree[num].maxCol = 0;
 75         return;
 76     }
 77     int mid = (Tree[num].left + Tree[num].right) >> 1;
 78     if(row <= mid)
 79         remove(row, col, 2 * num);
 80     else
 81         remove(row, col, 2 * num + 1);
 82     Tree[num].maxCol = myMax(Tree[2 * num].maxCol, Tree[2 * num + 1].maxCol);
 83 }
 84 
 85 int find(int row, int col, int num)
 86 {
 87     if(Tree[num].maxCol <= col)
 88         return -1;
 89     if(Tree[num].left == Tree[num].right)
 90     {
 91         if (Coord[Pos[Tree[num].left]].row > Coord[Pos[row]].row)
 92             return num;
 93         else
 94             return -1;
 95     }
 96 
 97     int mid = (Tree[num].left + Tree[num].right) >> 1;
 98     int res;
 99     if(mid > row)
100     {
101         res = find(row, col, 2 * num);
102         if(res != -1)
103             return res;
104     }
105     if(Tree[num].right > row)
106     {
107         res = find(row, col, 2 * num + 1);
108         if(res != -1)
109             return res;
110     }
111     return -1;
112 }
113 
114 int main()
115 {
116     freopen("in.txt","r",stdin);
117     int N;
118     int T = 1;
119     while(scanf("%d", &N) == 1 && N)
120     {
121         if(T != 1)
122             printf("\n");
123         printf("Case %d:\n", T++);
124         for(int i = 0; i < N; ++i)
125         {
126             scanf("%s%d%d", OP[i], &Coord[i].row, &Coord[i].col);
127             Pos[i] = i;
128         }
129         sort(Pos, Pos + N, comp);
130         Hash[Pos[0]] = 0;
131         int flag = 0;
132         for(int i = 1; i < N; ++i)
133         {
134             if(Coord[Pos[i]] == Coord[Pos[i - 1]])
135                 Hash[Pos[i]] = flag;
136             else
137             {
138                 Hash[Pos[i]] = ++flag;
139                 Pos[flag] = Pos[i];
140             }
141         }
142         build(0, flag, 1);
143         for(int i = 0; i < N; ++i)
144         {
145             if(strcmp(OP[i], "add") == 0)
146                 insert(Hash[i], Coord[i].col, 1);
147             else if(strcmp(OP[i], "remove") == 0)
148                 remove(Hash[i], Coord[i].col, 1);
149             else
150             {
151                 int ans = find(Hash[i], Coord[i].col, 1);
152                 if(ans != -1)
153                     printf("%d %d\n", Coord[Pos[Tree[ans].left]].row, Tree[ans].maxCol);
154                 else
155                     printf("-1\n");
156             }
157         }
158     }
159     return 0;
160 }