POJ1065——Wooden Sticks(动态规划，二分优化)

原题如下：

Wooden Sticks

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21590 Accepted: 9197

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

213

思路：遇到过好几个这样的题目了。就是把序列分为x个不下降子序列，求x的最小值。

一个定理是x就等于最长下降子序列的长度。不会证明QAQ。

直接dp就好。

#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;#define MAXN 5010#define INF 1e9+7#define MODE 1000000#define LIMIT 100000000000000000typedef long long ll;int t,n;struct woo{    int l;    int w;}a[MAXN];int cmp(woo a,woo b){    if(a.l!=b.l)        return a.l<b.l;    else        return a.w<b.w;}int dp[MAXN];//最长上升子序列的计算int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d%d",&a[i].l,&a[i].w);        }        sort(a,a+n,cmp);        int maxn=-1;        for(int i=0;i<n;i++){            dp[i]=1;            for(int j=0;j<i;j++)            {                if(a[j].w>a[i].w)                    dp[i]=max(dp[i],dp[j]+1);            }            maxn=max(maxn,dp[i]);        }        printf("%d\n",maxn);    }}

但这样不是最快的。因为求最长上升子序列和最长下降子序列的时候，最多有效更新只有一次。

用二分找到需要更新的位置即可、

代码如下。

#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;#define MAXN 5010#define INF 1e9+7#define MODE 1000000#define LIMIT 100000000000000000typedef long long ll;int t,n;struct woo{    int l;    int w;}a[MAXN];int cmp(woo a,woo b){    if(a.l!=b.l)        return a.l>b.l;    else        return a.w>b.w;}int dp[MAXN];//最长上升子序列的计算int main(){    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d%d",&a[i].l,&a[i].w);        }        sort(a,a+n,cmp);        fill(dp,dp+n,INF);        for(int i=0;i<n;i++)        {            *lower_bound(dp,dp+n,a[i].w)=a[i].w;        }        printf("%d\n",lower_bound(dp,dp+n,INF)-dp);    }}