Wooden Sticks

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 ) . 

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. 

The output should contain the minimum setup time in minutes, one per line. 

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 

213

Taejon 2001

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和poj3636同样的道理

因为偏序关系是<=，所以w从小到大相同l小的在前，找最长下降子序列

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int N=5005,INF=1e9;struct data{    int w,l;}da[N];bool cmpda(data a,data b){    if(a.w>b.w) return 0;    if(a.w<b.w) return 1;    if(a.w==b.w) return a.l<b.l?1:0;    return 1;}int t,n;int f[N],g[N],a[N];bool cmp(int a,int b){    return a>b;}int dp(){    int ans=0;    sort(da+1,da+1+n,cmpda);    memset(f,0,sizeof(f));    for(int i=1;i<=n;i++) g[i]=-INF,a[i]=da[i].l;    for(int i=1;i<=n;i++){        int k=lower_bound(g+1,g+1+n,a[i],cmp)-g;        f[i]=k;        g[k]=a[i];        ans=max(ans,f[i]);    }    return ans;}int main(int argc, const char * argv[]) {    scanf("%d",&t);    for(int i=1;i<=t;i++){        scanf("%d",&n);        for(int i=1;i<=n;i++) scanf("%d%d",&da[i].l,&da[i].w);        printf("%d\n",dp());    }    return 0;}