http://acm.hdu.edu.cn/showproblem.php?pid=1051
Wooden Sticks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24757 Accepted Submission(s): 9973
(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).
题目大意:
有一堆n个木棍,长度质量已知,机器处理木棍需要设置时间,规定
(1)第一根木棍的设置时间是1min
(2)前一个处理的木棍长度和质量小于等于后一个就不用设置时间,否则需要1min设置
找到最小建立时间。
如 给出(4,9)(5,2)(2,1)(3,5)(1,4)则最小建立时间(1,4)(3,5)(4,9)(2,1)(5,2)。
(本来想找个贪心的水题耍一下..结果发现完全看不出这道题的贪心解法...然后想到之前做的一个导弹拦截的问题..感觉二者有异曲同工之妙..
题目分析:首先按照长度进行排序..长度相同的按照重量排序..然后就能直接取了??(假的贪心..很容易找到hack数据..我的做法是在排序之后..从后往前找到这段数字的最长上升子序列的长度K,..则.K就是答案..这一点和导弹拦截的那题几乎一致。
导弹拦截http://acm.hdu.edu.cn/showproblem.php?pid=1257
#include <bits/stdc++.h> using namespace std; struct pot{
int len;
int weig;
}qwq[];
bool cmp(struct pot aa,struct pot bb){
if(aa.len!=bb.len)
return aa.len < bb.len;
return aa.weig < bb.weig;
}
int main()
{
int nums[];
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
memset(nums,,sizeof(nums));
for(int i = ; i < n; i++){
scanf("%d%d",&qwq[i].len,&qwq[i].weig);
}
sort(qwq,qwq+n,cmp);
int dp[];
int tot=;
dp[]=qwq[n-].weig;
int maxx=qwq[n-].weig;
for(int i = n- ; i >= ; i--){
if(qwq[i].weig>dp[tot]){
dp[++tot]=qwq[i].weig;
maxx=dp[tot];
}
else{
int l=;
int r=tot;
while(l<=r){
int mid=(l+r)/;
if(dp[mid]>=qwq[i].weig){
r=mid-;
}
else {
l=mid+;
} }
dp[l]=qwq[i].weig;
}
}
cout << tot <<endl;
}
return ;
}
(貌似导弹拦截那题大多数人也是使用贪心..(> _ <)