hdu 1051 Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25710    Accepted Submission(s): 10420

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)

 

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把给的木棒按照顺序排个队，如果当前木棒比前一个的长和重都不小，那么就不需要setup time，否则需要一分钟的setup time。拍个队，要求让setup time最小。实际上就是排成几个队，每个队都满足l[i + 1] >= l[i] && w[i + 1] >= w[i]。把这些队首尾相接即可，耗费的时间就是队的数量。贪心算法：先排序，从头开始找一个未标记的，往后找不耗时间的最近的木棒，依次标记。

 

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define inf 0x3f3f3f3f
#define MAX 5000
using namespace std;
typedef pair<int,int> pa;
int t,n;
pa p[MAX];
bool vis[MAX];
void dfs(int k) {
    vis[k] = true;
    for(int i = k + 1;i < n;i ++) {
        if(vis[i]) continue;
        if(p[i].first >= p[k].first && p[i].second >= p[k].second) {
            dfs(i);
            break;
        }
    }
}
int main() {
    scanf("%d",&t);
    while(t --) {
        scanf("%d",&n);
        for(int i = 0;i < n;i ++) {
            vis[i] = false;
            scanf("%d%d",&p[i].first,&p[i].second);
        }
        sort(p,p + n);
        int ans = 0;
        for(int i = 0;i < n;i ++) {
            if(!vis[i]) {
                dfs(i);
                ans ++;
            }
        }
        printf("%d
",ans);
    }
}