codeforces4D - Mysterious Present DP

题意：求限定起始位置的二维最长递增子序列.

解题思路：直接DP

解题代码：

  1 // File Name: 4d.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年08月04日 星期一 19时24分49秒
  4
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25
 26 using namespace std;
 27
 28 struct node{
 29   int from;
 30   int len;
 31 }dp[5000];
 32 struct node1{
 33   int w, h,s;
 34 }a[5005];
 35 int ok[10000];
 36 int cmp(node1 a,node1 b)
 37 {
 38    if(a.w == b.w)
 39        return a.h < b.h;
 40    return a.w < b.w;
 41 }
 42 void dfs(int k)
 43 {
 44   if(k == -1)
 45       return;
 46    dfs(dp[k].from);
 47    if(a[k].s)
 48    printf("%d ",a[k].s);
 49
 50 }
 51 int main(){
 52    int n ;
 53    scanf("%d %d %d",&n,&a[0].w,&a[0].h);
 54    memset(ok,0,sizeof(ok));
 55    for(int i =1 ;i <= n;i ++)
 56    {
 57        scanf("%d %d",&a[i].w,&a[i].h);
 58        a[i].s = i;
 59    }
 60    sort(a+1,a+1+n,cmp);
 61  /*  for(int i = 1;i <= n;i ++)
 62    {
 63        printf("%d %d\n",a[i].w,a[i].h);
 64    }*/
 65    for(int i = 0 ;i <= n;i ++)
 66    {
 67       dp[i].from = i ;
 68       dp[i].len = 1;
 69    }
 70    dp[0].from = -1;
 71    ok[0] = 1;
 72    int asite = 0;
 73    int amx = 1;
 74    for(int i = 1;i <= n;i ++)
 75    {
 76       int mx = 0;
 77       int site = -1 ;
 78       for(int j = 0;j < i;j ++)
 79       {
 80         if(a[i].w > a[j].w && a[i].h > a[j].h && ok[j])
 81         {
 82             ok[i] = 1;
 83             if(dp[j].len > mx)
 84             {
 85                mx = dp[j].len;
 86                site = j;
 87             }
 88         }
 89       }
 90       dp[i].len = mx +1;
 91       dp[i].from = site;
 92       if(dp[i].len > amx)
 93       {
 94         amx = dp[i].len ;
 95         asite = i;
 96       }
 97       //printf("%d\n",dp[i].len);
 98    }
 99    printf("%d\n",amx-1);
100    dfs(asite);
101 return 0;
102 }

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