HDU 1051: Wooden Sticks(贪心)
Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11244 Accepted Submission(s): 4627
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).
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.
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
题意是第一个3是3组数据, 然后一个5是5根棍子。 然后逐一输入其的长度和重量。
。题目是问怎么做这些棍子用时最少,在做棍子时。若后一根棍子。长度和重量都大于前
一根。则不用加时间。。
想法。。
拿第一组数据来说。
。
首先按长度以小到大排序:(1,4),(2,1),(3。 5)。 (4, 9),(5,2);
我们发现(2。 1)的重量太小, 而(5, 2)的长度太长。这样排结果为3;
事实上这时,我们仅仅要把不满足的先标记起来。
。
则(1,4)。(3,5),(4,9) ———— (2,1), (5,2)
但我们这么排, 也就是题目给出的顺序,结果就为2;也就是最优。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath> using namespace std; #define M 100500 struct node
{
int lenth;
int weight;
int k;
} Q[M]; bool cmp ( node a, node b )
{
if(a.lenth==b.lenth)
return a.weight<b.weight;
else
return a.lenth<b.lenth;
} int main()
{
int t;
int n;
scanf("%d", &t);
while( t-- )
{
scanf("%d", &n);
for(int i=0; i<n; i++)
{
scanf("%d%d", &Q[i].lenth, &Q[i].weight);
Q[i].k= 1;
}
sort( Q, Q+n, cmp );
int ans = 0;
for(int i=0; i<n-1; i++)
{
if(!Q[i].k)
continue;
int t = Q[i].weight;
for(int j = i+1; j<n; j++)
if(Q[j].weight>=t &&Q[j].k==1 )
{
ans++;
Q[j].k=0;
t = Q[j].weight;
}
}
printf("%d\n", n-ans);
} return 0;
}
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