ZOJ1025 Wooden Sticks

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

Output for the Sample Input

2
1
3

思路：

题意大概是锯木桩，每次小于前面开过机的长和厚不用建立，大于的话就得重新发动，发动时间为1，为求最小发动时间。其实就是求最少组的非递减子序列。

第一种动态规划吧，相当于是最长不下降子序列长度问题。嗯，先对l.w升序排，先排长度，长度相等时在按重量的升序排。再求重量的最长下降子序列长度L，即为所求。然而我并不会证。

第二种贪心吧，一边找一边标记找过了。

代码：

DP

#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<vector>
using namespace std;
struct node
{
    int l, w;
};
bool cmp(const node& a,const node& b){
    if (a.l == b.l){
        return a.w < b.w;
    }
    return a.l < b.l;
}
node sticks[5005];
int main()
{
    int T,n;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        memset(sticks,0,sizeof(sticks));
        for (int i = 0; i < n; i++){
            scanf("%d%d",&sticks[i].l,&sticks[i].w);
        }
        sort(sticks,sticks+n,cmp);
        vector<int> dp(n, 1);
        int longest = 1;    
        for (int i = 1; i < n; i++){
            for (int j = 0; j < i; j++){
                if (sticks[i].w < sticks[j].w){
                    dp[i] = max(dp[i],dp[j]+1);
                }
            }
            longest = max(longest,dp[i]);
        }
        printf("%d\n",longest);
    }
    return 0;
}

贪心：

#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
struct node{
    int l, w;
    bool flag;
    node() :l(0), w(0), flag(false){

    }
};
bool cmp(const node& a,const node& b){
    if (a.l == b.l){
        return a.w < b.w;
    }
    return a.l < b.l;
}
int main()
{
    int T, n;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        node* sticks = new node[n];
        for (int i = 0; i < n;i++){
            scanf("%d%d",&sticks[i].l,&sticks[i].w);
        }
        sort(sticks,sticks+n,cmp);
        int res=0;
        for (int i = 0; i < n; i++){
            if (sticks[i].flag){
                continue;
            }
            res++;
            int tmp = sticks[i].w;
            for (int j = i; j < n; j++){
                if (tmp <= sticks[j].w && !sticks[j].flag){
                    sticks[j].flag = true;
                    tmp = sticks[j].w;
                }
            }
        }
        printf("%d\n",res);
    }
    return 0;
}