原題鏈接 Wooden Sticks Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 22711 Accepted: 9731 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 Source
Taejon 2001 思路：題目的確需要稍加思考，這道題的要求其實是將所有stick分為x個不下降子序列( Ai <= Ai+1 ),然后問題歸結于求x的最小值。 x的最小值其實等于按l遞增排序后stick按w最長下降子序列的長度L，證明如下： 若x < L，先從stick中取出最長下降子序列L，取走的元素留下一個大小相同的“空穴”， 然后將剩下的元素和空穴分成x個不下降子序列。接著把最長下降子序列L中的L個元素放回這L個空穴里。 由于x < L，所以根據鴿籠原理，必然有兩個或兩個以上的下降子序列L中的元素(b > a)被按順序放到同一個不下降子序列(a <= b)， 而這x個序列本應該是不下降的，這就產生了矛盾，所以x >= L,所以x的最小值就是L了。
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5000 + 10;typedef struct sticks{ int l,w;}sticks;sticks a[maxn];int dp[maxn],n;//dp[i+1]：最長下降子序列第i位的數字大小bool cmp(sticks x,sticks y){ if(x.l < y.l) return true; else if(x.l==y.l) return x.w<y.w ? true : false; else return false;}//從一個遞減序列中返回第一個>=x的數的下標int serch(int x){ int l=-1,r=maxn-1,mid; while(r-l>1){ mid=(l+r)/2; if(dp[mid]>x) l=mid; else r=mid; } return r;}int main(){ int T; cin >> T; while(T--){ cin >> n; for(int i=0;i<n;i++) scanf("%d%d",&a[i].w,&a[i].l); sort(a,a+n,cmp); memset(dp,-1,sizeof(dp)); for(int i=0;i<n;i++) dp[serch(a[i].w)] = a[i].w; cout << serch(-1) << endl; }}新聞熱點
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