ZOJ1025 Wooden Sticks
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题目:
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
Output for the Sample Input
2
1
3
思路:
题意大概是锯木桩,每次小于前面开过机的长和厚不用建立,大于的话就得重新发动,发动时间为1,为求最小发动时间。其实就是求最少组的非递减子序列。
第一种动态规划吧,相当于是最长不下降子序列长度问题。嗯,先对l.w升序排,先排长度,长度相等时在按重量的升序排。再求重量的最长下降子序列长度L,即为所求。然而我并不会证。
第二种贪心吧,一边找一边标记找过了。
代码:
DP
#include<iostream>#include<algorithm>#include<string.h>#include<stdio.h>#include<stdlib.h>#include<vector>using namespace std;struct node{ int l, w;};bool cmp(const node& a,const node& b){ if (a.l == b.l){ return a.w < b.w; } return a.l < b.l;}node sticks[5005];int main(){ int T,n; scanf("%d",&T); while (T--){ scanf("%d",&n); memset(sticks,0,sizeof(sticks)); for (int i = 0; i < n; i++){ scanf("%d%d",&sticks[i].l,&sticks[i].w); } sort(sticks,sticks+n,cmp); vector<int> dp(n, 1); int longest = 1; for (int i = 1; i < n; i++){ for (int j = 0; j < i; j++){ if (sticks[i].w < sticks[j].w){ dp[i] = max(dp[i],dp[j]+1); } } longest = max(longest,dp[i]); } printf("%d\n",longest); } return 0;}
贪心:
#include<iostream>#include<string.h>#include<algorithm>#include<stdio.h>#include<stdlib.h>using namespace std;struct node{ int l, w; bool flag; node() :l(0), w(0), flag(false){ }};bool cmp(const node& a,const node& b){ if (a.l == b.l){ return a.w < b.w; } return a.l < b.l;}int main(){ int T, n; scanf("%d",&T); while (T--){ scanf("%d",&n); node* sticks = new node[n]; for (int i = 0; i < n;i++){ scanf("%d%d",&sticks[i].l,&sticks[i].w); } sort(sticks,sticks+n,cmp); int res=0; for (int i = 0; i < n; i++){ if (sticks[i].flag){ continue; } res++; int tmp = sticks[i].w; for (int j = i; j < n; j++){ if (tmp <= sticks[j].w && !sticks[j].flag){ sticks[j].flag = true; tmp = sticks[j].w; } } } printf("%d\n",res); } return 0;}
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