hdu1051 Wooden Sticks(贪心)
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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .
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
题意是给定根木棍,以及每根木棍的长度和重量。现在要对它们进行加工:如果加工完一根长重的木棍后,下一根长重的木棍满足且,那么机器就可以继续加工而不需要进入调整时间。否则的话,需要1分钟的调整时间才能继续加工。求总共最小需要的调整时间。
贪心:双关键字排序,然后寻找满足且的序列个数。
#include<cstdio>
#include<algorithm>
#define N 5010
struct node{
int l, w, vis;
bool operator<(const node&rhs) const
{
return l==rhs.l?w<rhs.w:l<rhs.l;
}
}stick[N];
int n;
void init() {
memset(&stick,0,sizeof stick);
scanf("%d", &n);
for(int i=0; i<n; i++)
scanf("%d%d", &stick[i].l, &stick[i].w);
}
void find(int i) {
int Lasti = i;
for(int j=i+1; j<n; j++)
if(!stick[j].vis && stick[Lasti].w <= stick[j].w)
{
stick[j].vis = 1;
Lasti=j;
}
}
void solve() {
int res = 0;
std::sort(stick, stick+n);
for(int i=0; i<n; i++)
if(!stick[i].vis)
{
stick[i].vis = 1;
++res;
find(i);
}
printf("%d\n", res);
}
int main() {
int T;
scanf("%d", &T);
while(T--)
{
init();
solve();
}
return 0;
}
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