poj 1065 Wooden Sticks

题目大意：有一堆木棍，属性有长度和重量。它们需要被处理，若后面的长度和重量都大于前面的，消耗为0，否则为1。第一个消耗为1，求按一定的顺序处理，最小消耗是多少？

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 2n 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

这个题我们先按长度排序，接着去寻找重量的不下降子序列的个数，这个个数就是消耗值
代码如下

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef pair<int ,int> Stick;//length, weight
Stick stick[10002];
int flag[10002];

int cmp(Stick a, Stick b) {
    if(a.first == b.first) {
        return a.second < b.second;
    }
    return a.first < b.first;
}

int t,n;

int main(int argc, char const *argv[])
{
    //freopen("input.txt","r",stdin);
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(int i = 0; i < n; i++) {
            scanf("%d %d",&stick[i].first,&stick[i].second);
        }
        sort(stick, stick+n, cmp);
        
        memset(flag, 0, sizeof(flag));
        
        bool ok = false;
        int cost = 0;
        
        while(!ok) {
            int last = -1;
            ok = true;
            for(int i = 0; i < n; i++) {
                if(!flag[i] && (last == -1 || stick[i].second >= stick[last].second)) {
                    flag[i] = 1;
                    last = i;
                    ok = false;
                }
            }
            if(ok == false) {
                cost++;
            }
        }
        printf("%d
", cost);
        
    }

    return 0;
}