zoukankan      html  css  js  c++  java
  • Ping-Pong (Easy Version)(DFS)

    B. Ping-Pong (Easy Version)
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to interval (c, d) from our set if and only if c < a < d or c < b < d. Also there is a path from interval I1 from our set to interval I2 from our set if there is a sequence of successive moves starting from I1 so that we can reach I2.

    Your program should handle the queries of the following two types:

    1. "1 x y(x < y) — add the new interval (x, y) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.
    • "2 a b(a ≠ b) — answer the question: is there a path from a-th (one-based) added interval to b-th (one-based) added interval?

    Answer all the queries. Note, that initially you have an empty set of intervals.

    Input

    The first line of the input contains integer n denoting the number of queries, (1 ≤ n ≤ 100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.

    It's guaranteed that all queries are correct.

    Output

    For each query of the second type print "YES" or "NO" on a separate line depending on the answer.

    近来很颓。。。

     1 #include <iostream>
     2 #include <string>
     3 #include <algorithm>
     4 #include <map>
     5 #include <vector>
     6 #include <cstdio>
     7 #include <cmath>
     8 #include <cstring>
     9 using namespace std;
    10 
    11 const int MAXN = 102;
    12 int a[MAXN], b[MAXN];
    13 bool f[MAXN];
    14 int n, cnt;
    15 
    16 void dfs(int s, int e)
    17 {
    18     f[s] = 1;
    19     if(f[e]) return ;
    20     for(int i = 1; i < cnt; i++)
    21     {
    22         if(f[i]) continue;
    23         if(a[s] > a[i] && a[s] < b[i]) dfs(i, e);
    24         if(f[e]) return ;
    25         if(b[s] > a[i] && b[s] < b[i]) dfs(i, e);
    26         if(f[e]) return ;
    27     }
    28 }
    29 
    30 int main()
    31 {
    32     while(scanf("%d", &n) != EOF)
    33     {
    34         cnt = 1;
    35         int m, x, y;
    36         while(n--)
    37         {
    38             scanf("%d", &m);
    39             if(m == 1)
    40             {
    41                 scanf("%d %d", &a[cnt], &b[cnt]);
    42                 cnt++;
    43             }
    44             else
    45             {
    46                 scanf("%d %d", &x, &y);
    47                 memset(f, 0, sizeof(f));
    48                 dfs(x, y);
    49                 if(f[y]) puts("YES");
    50                 else puts("NO");
    51             }
    52         }
    53     }
    54     return 0;
    55 }
    View Code

  • 相关阅读:
    【转】《基于MFC的OpenGL编程》Part 5 Transformations Rotations, Translations and Scaling
    【转】 《基于MFC的OpenGL编程》Part 10 Texture Mapping
    【转】 《基于MFC的OpenGL编程》Part 11 Blending, Antialiasing and Fog
    win form 托盘功能的实现(引用CSDN)
    C# win form退出窗体时对话框实用
    智能DNS 笔记
    iis无法启动, 找出占用80端口的罪魁祸首
    gvim for windows的剪贴板操作
    内容交换
    Content Networking 读书笔记
  • 原文地址:https://www.cnblogs.com/cszlg/p/3244138.html
Copyright © 2011-2022 走看看