WoodenSticks(区间贪心类题目)
Description
There 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 (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is
a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).
Input
The 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 magnitude 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.
Output
The output should 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
2
1
3
这个题目是贪心类题目,算是做的第二个贪心类题目了。
思路就是先把w1或者l1其中的一个从小到大排列。我是把w从小到大排列的,但是这其中可能有w相同时l也应该从小到大排列。
排完之后刚开始我只是找了一组1分钟的,意思是其他满足上述条件的一组就是一分钟了。所以那些测试数据都也能通过,但是就是wrong answer,后来找到了原因改了代码终于对了。 好、还是请大神们指点。
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; int main() { int a; int x[10001][2]; scanf("%d",&a); while(a--) { int b; scanf("%d",&b); for (int number=0;number<b;number++) { scanf("%d %d",&x[number][0],&x[number][1]); } int Mix,temp; for (int number1=0;number1<b;number1++) { Mix=number1; for (int number2=number1+1;number2<b;number2++) { if (x[Mix][1]>x[number2][1]) Mix=number2; } if (Mix!=number1) { temp=x[Mix][1]; x[Mix][1]=x[number1][1]; x[number1][1]=temp; temp=x[Mix][0]; x[Mix][0]=x[number1][0]; x[number1][0]=temp; } } for (int number=0;number<b-1;number++) { if (x[number][1]==x[number+1][1]) { if (x[number][0]>x[number+1][0]) { temp=x[number][0]; x[number][0]=x[number+1][0]; x[number+1][0]=temp; } } } int Count=0,Max; for (int number1=0;number1<b;number1++) { if (x[number1][0]!=-1) { Count++; Max=x[number1][0]; for (int number2=number1+1;number2<b;number2++) { if (x[number2][0]!=-1) { if (x[number2][0]>=Max) { Max=x[number2][0]; x[number2][0]=-1; } } } } } printf("%d\n",Count); } return 0; }