1410260946-hd-wooden sticks

Wooden sticks

Time Limit: 2000/1000 ms (Java/other) memory limit: 65536/32768 K (Java/other) total submission (s): 1 accepted submission (s): 1 Font: times New Roman | verdana | georgiafont size: Batch → problem descriptionthere is a pile of N wooden sticks. the length and weight of each stick are known in advance. the sticks are to be processed by a woodworking machine in one by one fashion. it needs some time, called setup time, for the machine to prepare processing a stick. the setup times are associated with cleaning operations and changing tools and shapes in the machine. the setup times of the woodworking machine are given as follows:

(A) The setup time for the first wooden stick is 1 minute.
(B) Right after processing a stick of length L and weight W, the machine will need no setup time for a stick of length l' and weight W' if l <= l' and W <= W '. otherwise, it will need 1 minute for setup.

You are to find the minimum setup time to process a given pile of N wooden sticks. for example, if you have five sticks whose pairs of length and weight are (), and ), then the minimum setup time shocould be 2 minutes since there is a sequence of pairs ).
Inputthe input consists of T test cases. the number of test cases (t) is given in the first line of the input file. each test case consists of two lines: the first line has an integer N, 1 <= n <= 5000, that represents the number of wooden sticks in the test case, and the second line contains N 2 positive integers L1, W1, L2, W2 ,..., ln, Wn, each of magnloud at most 10000, where Li and WI are the length and weight of the I th wooden stick, respectively. the 2n integers are delimited by one or more spaces.
Outputthe output shoshould contain the minimum setup time in minutes, one per line.
Sample Input

3 5 4 9 5 2 2 1 3 5 1 4 3 2 2 1 1 2 2 3 1 3 2 2 3 1

Sample output

213

Theme

Processing a pile of wood, length and weight do not want to wait, if the length and weight of one of the smaller, You need to manipulate the machine, each operation takes one minute, at first need to manipulate once. Calculate the minimum time.

Solutions

Based on the meaning of the question, two cross sorting is required. To solve this problem, you can sort a condition in a quick order, and then consider the second condition in particular.

Code

# Include <stdio. h ># include <algorithm> using namespace STD; struct wood {int L, t;} Woods [5050]; bool CMP (wood A, wood B) {if (. l! = B. l) return. L <B. l; else return. T <B. t;} int main () {int t, n; int I, j, sum; int A; scanf ("% d", & T); While (t --) {scanf ("% d", & N); for (I = 0; I <n; I ++) scanf ("% d ", & Woods [I]. l, & Woods [I]. t); sort (woods, Woods + N, CMP); // sort a condition first by sum = 1; A = Woods [0]. t; Woods [0]. T = 0; // mark it as while (1) {for (I = 1; I <n; I ++) if (a <= Woods [I]. t) // error // when comparing the size, pay special attention to how to judge {A = Woods [I] When the values are equal. t; Woods [I]. T = 0 ;}for (I = 0, j = 0; I <n; I ++) {If (Woods [I]. t! = 0) {A = Woods [I]. t; break;} else J ++;} // If (j = N) // when the number of tags is equal to the total number, the break is displayed. Else sum ++;} printf ("% d \ n", sum);} return 0 ;}

 

1410260946-hd-wooden sticks