HDU 1051:Wooden Sticks
Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11605 Accepted Submission(s): 4792
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).
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.
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
2
1
3
简单贪心。。
题意能够看做是: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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