zoukankan      html  css  js  c++  java
  • uva.10020 Minimal coverage(贪心)

    10020


    Given several segments of line (int the X axis) with coordinates [Li
    , Ri
    ]. You are to choose the minimal
    amount of them, such they would completely cover the segment [0, M].
    Input
    The first line is the number of test cases, followed by a blank line.
    Each test case in the input should contains an integer M (1 ≤ M ≤ 5000), followed by pairs “Li Ri”
    (|Li
    |, |Ri
    | ≤ 50000, i ≤ 100000), each on a separate line. Each test case of input is terminated by pair
    ‘0 0’.
    Each test case will be separated by a single line.
    Output
    For each test case, in the first line of output your programm should print the minimal number of line
    segments which can cover segment [0, M]. In the following lines, the coordinates of segments, sorted
    by their left end (Li), should be printed in the same format as in the input. Pair ‘0 0’ should not be
    printed. If [0, M] can not be covered by given line segments, your programm should print ‘0’ (without
    quotes).
    Print a blank line between the outputs for two consecutive test cases.
    Sample Input
    2
    1
    -1 0
    -5 -3
    2 5
    0 0
    1
    -1 0
    0 1
    0 0
    Sample Output
    0
    1
    0 1

    贪心是要讲规则的,这道题的规则就是:首先要覆盖x=0这个点,那么选谁呢?自然选L<=0 && R 尽可能大的区间[l0,r0];然后下一个要覆盖的是x=ro,采取同样的策略……直到某一轮的选出的区间覆盖了x=m。

     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<algorithm>
     4 const int inf = 0x3f3f3f3f ;
     5 int T ;
     6 int m ;
     7 int path[100000 + 10] ;
     8 struct node
     9 {
    10     int l , r ;
    11 }e[100000 + 10];
    12 
    13 bool cmp (node a , node b)
    14 {
    15     return a.l < b.l ;
    16 }
    17 
    18 int main ()
    19 {
    20     //freopen ("a.txt" , "r" , stdin ) ;
    21     scanf ("%d" , &T) ;
    22     int cas = 0 ;
    23     while (T --) {
    24         if (cas != 0 ) puts ("") ;
    25         cas ++ ;
    26         scanf ("%d" , &m ) ;
    27         int n = 0 ;
    28         while (1) {
    29             scanf ("%d%d" , &e[n].l , &e[n].r ) ;
    30             if (e[n].l == 0 && e[n].r == 0) break ;
    31             n ++ ;
    32         }
    33        //printf ("n = %d
    " , n );
    34         std::sort (e , e + n , cmp) ;
    35         bool flag = 0 ;
    36         int cur = 0 , id = -1 , maxn = -inf ;
    37         int tot = 0 ;
    38         for (int i = 0 ; i < n ; i ++) {
    39             for (int j = 0 ; j < n ; j ++) {
    40                // printf ("l = %d , r = %d , cur = %d
    " , e[j].l , e[j].r , cur ) ;
    41                 if (e[j].r > cur && e[j].l <= cur ) {
    42                     if (e[j].r > maxn) {
    43                         id = j ;
    44                         maxn = e[j].r ;
    45                     }
    46                 }
    47             }
    48             if (id == -1) {
    49                     flag = 1 ;
    50                     break ;
    51             }
    52             path[tot ++] = id ;
    53             cur = e[id].r ;
    54             id = -1 ; maxn = -inf ;
    55             if (cur >= m) break ;
    56         }
    57         if (flag) puts ("0") ;
    58         else {
    59             printf ("%d
    " , tot ) ;
    60             for (int i = 0 ; i < tot ; i ++) {
    61                 printf ("%d %d
    " , e[path[i]].l , e[path[i]].r ) ;
    62             }
    63         }
    64     }
    65     return 0 ;
    66 }
    View Code
  • 相关阅读:
    Struts2框架
    读者写者问题
    哲学家就餐问题
    理解中断
    理解处理机调度
    理解死锁
    理解进程
    Linux CentOS 6.7 挂载U盘
    家庭-养老院模型理解IOC和DI
    Bash基础
  • 原文地址:https://www.cnblogs.com/get-an-AC-everyday/p/4515133.html
Copyright © 2011-2022 走看看