关于最长上升子序列的链接：
http://wenku.baidu.com/view/fe0deecea1c7aa00b52acb71.html
http://blog.sina.com.cn/s/blog_4b1e4fe9010098af.html
http://www.cnblogs.com/celia01/archive/2012/07/27/2611043.html
裸的最长上升子序列：时间复杂度为O（n*n）
POJ 2533
贴代码：
1 #include<cstdio> 2 #define N 1005 3 int a[N]; 4 int lis(int n) 5 { 6 int *dp= new int[n+1]; 7 dp[1] =1; 8 int ans =1; 9 for(int i=2; i<=n; ++i) 10 { 11 int mx =0; 12 for(int j=1; j<i; ++j) 13 if(a[i] > a[j] && dp[j] > mx) 14 mx = dp[j]; 15 dp[i] = mx+1; 16 if(dp[i] > ans) ans = dp[i]; 17 } 18 return ans; 19 } 20 int main() 21 { 22 // freopen("in.c","r",stdin); 23 int n; 24 scanf("%d",&n); 25 for(int i=1; i<=n; ++i) 26 scanf("%d",&a[i]); 27 printf("%d ",lis(n)); 28 }
POJ 1631 求题意，反正大家都说是求最长上升子序列。
贴代码：复杂度，O（nlogn）,超短的代码长度
1 #include<cstdio> 2 #include<algorithm> 3 #define INF 0x3f3f3f3f 4 int dp[40005]; 5 using namespace std; 6 int main() 7 { 8 // freopen("in.c","r",stdin); 9 int t; 10 scanf("%d",&t); 11 while(t--) 12 { 13 int n,a; 14 scanf("%d",&n); 15 fill(dp,dp+n,INF); 16 for(int i=0; i<n; ++i) 17 { 18 scanf("%d",&a); 19 *lower_bound(dp,dp+n,a) = a; 20 } 21 printf("%d ",lower_bound(dp,dp+n,INF)-dp); 22 } 23 return 0; 24 }
手动写二分的代码：
1 #include <cstdio> 2 int D[40005]; 3 int bs(int low,int high,int m) 4 { 5 int mid = (low+high)/2; 6 while(low<=high) 7 { 8 if(D[mid] < m && D[mid+1] >= m) return mid; 9 else if(D[mid] < m) low = mid+1; 10 else high = mid-1; 11 mid = (low+high)/2; 12 } 13 return mid; 14 } 15 int main() 16 { 17 // freopen("in.c","r",stdin); 18 int t; 19 scanf("%d",&t); 20 while(t--) 21 { 22 int n,a; 23 scanf("%d",&n); 24 scanf("%d",&a); 25 D[1] = a; 26 int len =1; 27 for(int i=1; i<n; ++i) 28 { 29 scanf("%d",&a); 30 if(a > D[len]) 31 D[++len] = a; 32 else 33 { 34 int j=bs(1,len,a); 35 D[j+1] = a; 36 } 37 } 38 printf("%d ",len); 39 } 40 return 0; 41 }
codeforces 4D：
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29608#problem/A
只留下那些宽和高都严格大于卡片的信封，如果一张都没剩下，输出0，否则，先将卡片按width的值顺序，hight的值逆序排。然后按照高度值求一次最长上升子序列，记录路径即可···
贴代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 #define N 5005 6 int path[N]; 7 struct card 8 { 9 int w,h,ind; 10 } p[N]; 11 int cmp(card a,card b) 12 { 13 if(a.w == b.w) return a.h > b.h; 14 return a.w<b.w; 15 } 16 int lcs(int n) 17 { 18 int *dp = new int[n+1]; 19 dp[1] = 1; 20 int ans = 1; 21 int e =1; 22 for(int i=2; i<=n; ++i) 23 { 24 int mx =0,v=0; 25 for(int j=1; j<i; ++j) 26 { 27 if(dp[j] > mx && p[i].h > p[j].h) 28 { 29 mx = dp[j]; 30 v = j; 31 } 32 } 33 dp[i] = mx+1; 34 path[i] =v; 35 if(dp[i] > ans) 36 { 37 ans = dp[i]; 38 e = i; 39 } 40 } 41 printf("%d ",ans); 42 return e; 43 } 44 void dfs(int x) 45 { 46 if(!x) return; 47 dfs(path[x]); 48 if(path[x]) printf(" "); 49 printf("%d",p[x].ind); 50 } 51 int main() 52 { 53 // freopen("in.c","r",stdin); 54 int n,cnt=0,w0,h0,w,h; 55 scanf("%d%d%d",&n,&w0,&h0); 56 for(int i=1; i<=n; ++i) 57 { 58 int w,h; 59 scanf("%d%d",&w,&h); 60 if(w > w0 && h > h0) 61 { 62 ++cnt; 63 p[cnt].w = w,p[cnt].h = h,p[cnt].ind =i; 64 } 65 } 66 sort(p+1,p+cnt+1,cmp); 67 if(!cnt) printf("0 "); 68 else 69 { 70 int e = lcs(cnt); 71 dfs(e); 72 } 73 return 0; 74 }
POJ 1887 最长不降子序列
贴代码：
1 #include <cstdio> 2 int D[40005]; 3 int bs(int low,int high,int m) 4 { 5 int mid = (low+high)/2; 6 while(low<=high) 7 { 8 if(D[mid] > m && D[mid+1] <= m) return mid; 9 else if(D[mid] > m) low = mid+1; 10 else high = mid-1; 11 mid = (low+high)/2; 12 } 13 return mid; 14 } 15 int main() 16 { 17 // freopen("in.c","r",stdin); 18 int a; 19 int cas=0; 20 while(true) 21 { 22 scanf("%d",&a); 23 if(a == -1) break; 24 D[1] = a; 25 int len =1; 26 while(true) 27 { 28 scanf("%d",&a); 29 if(a == -1) break; 30 if(a <= D[len]) 31 D[++len] = a; 32 else 33 { 34 int j=bs(1,len,a); 35 D[j+1] = a; 36 } 37 } 38 printf("Test #%d: maximum possible interceptions: %d ",++cas,len); 39 } 40 return 0; 41 }