Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11249 Accepted Submission(s): 4629

Problem 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

Source Asia 2001, Taejon (South Korea)
Recommend We have carefully selected several similar problems for you: 1050 2037 1052 1045 1009

在POJ呆惯了，WA一直以为是代码问题，没想到是数组开小了。。。。
这道题关键在于为什么贪心可以找到最优解。
刚开始我不懂，为什么用贪心可以找出最优解！也在这个问题上纠结了很久，感觉比较痛苦！
这和找最长递增子序列不同，只要能加入到原有的序列中比新开一个递增子序列，所得到的子序列数一定更少。

#include 
  
   
#include 
    
    using namespace std
    ; #define M 11000 
    struct node
    { int l
    ,w
    ; }f
    [M
    ]; int vis
    [M
    ]; //将长度排序，降低成为一位数组的扫描。 bool cmp
    (node a
    ,node b
    ){ if(a
    .l
    <b
    .l
    ) return true
    ; if(a
    .l
    >b
    .l
    ) return false
    ; if(a
    .l
    ==b
    .l
    ) return a
    .w
    <b
    .w
    ; } int main() { int n
    ,m
    ,t
    ,i
    ,j
    ,cur
    ,tot
    ; while(scanf
    ("%d"
    ,&n
    )!=EOF
    ) { while(n
    --) { cur
    =0
    ;tot
    =0
    ; memset
    (vis
    ,0
    ,sizeof(vis
    )); scanf
    ("%d"
    ,&m
    ); for(i
    =0
    ;i
    <m
    ;i
    ++) { scanf
    ("%d%d"
    ,&f
    [i
    ].l
    ,&f
    [i
    ].w
    ); } sort
    (f
    ,f
    +m
    ,cmp
    ); for(i
    =0
    ;i
    <m
    ;i
    ++) { if(vis
    [i
    ]) continue; //如果已经排入一个递增子序列，就不用再考虑。 t
    =f
    [i
    ].w
    ;vis
    [i
    ]=1
    ; for(j
    =i
    +1
    ;j
    <m
    ;j
    ++) { if(!vis
    [j
    ]&&t
    <=f
    [j
    ].w
    ) //找递增子序列的元素。 { vis
    [j
    ]=1
    ; t
    =f
    [j
    ].w
    ; } } tot
    ++; } printf
    ("%d\n"
    ,tot
    ); } } return 0
    ; }

HDU 1051 Wooden Sticks (贪心)

Wooden Sticks