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HDU 1051：Wooden Sticks

2024-07-15 04:31:24   221人阅读

Wooden Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11605    Accepted Submission(s): 4792


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

简单贪心。。

题意可以看做是：给定若干（1<= n <=5000）组二维坐标点，凡是满足  "x1<= x2 && y1<= y2"的话那么我们承认这两个坐标是属于同一个集合中。题目要我们求出这些坐标点最少能表示成几个集合。





#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;

const int maxn = 5000 + 50;

int cas;//案例
int n;//木条数量
int vis[maxn];//标记数组
int ans_minute;//最短时间

struct point
{
    int l;//长度
    int w;//重量
};
point s[maxn];

bool cmp(point a, point b)
{
    if( a.l==b.l )return a.w<b.w;
    else  return a.l<b.l;
}

int main()
{
    scanf("%d", &cas);
    while( cas-- )
    {
        scanf("%d", &n);
        ans_minute = 0;
        memset(vis, 0, sizeof(vis));
        for(int i=1; i<=n; i++)scanf("%d%d", &s[i].l, &s[i].w);
        sort(s+1, s+n+1, cmp);
        //for(int i=1; i<=n; i++)printf("%d__%d  ", s[i].l, s[i].w);
        int judge = 1;
        for(int i=1; i<=n; i++)
        {
            if( vis[i] ) continue;
            ans_minute++;
            int temp = s[i].w;
            vis[i] = 1;
            for(int j=i+1; j<=n; j++)
            {
                if( temp<=s[j].w && !vis[j] )
                {
                    temp = s[j].w;
                    judge++;
                    vis[j] = 1;
                    continue;
                }
            }
            if(judge>=n)break;
        }

         printf("%d\n", ans_minute);
    }
}




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