Longest ascending subsequence LIS set POJ2533, POJ1631, POJ1887, POJ1609

POJ2533 finds the longest ascending subsequence naked, and the complexity is the template of n ^ 2.

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              # Define ll long # define eps 1e-7 # define inf 0 xfffffff // const ll INF = 1ll <61; using namespace std; // vector
             
               > G; // typedef pair
              
                P; // vector
               
                 >:: Iterator iter; // map
                
                  Mp; // map
                 
                   : Iterator p; // # define IN freopen ("c: \ Users \ linzuojun \ desktop \ input.txt", "r", stdin) // # define OUT freopen ("c: \ Users \ linzuojun \ desktop \ output.txt", "w", stdout) int dp [10000 + 5]; int num [10000 + 5]; void clear () {memset (dp, 0, sizeof (dp); memset (num, 0, sizeof (num ));} int main () {int n; while (cin> n) {memset (dp, 0, sizeof (dp); for (int I = 1; I <= n; I ++) cin> num [I]; int ans = 1; for (int I = 1; I <= n; I ++) dp [I] = 1; for (int I = 2; I <= n; I ++) {int m = 0; for (int j = 1; j
                  
                    Num [j] & dp [j] + 1> dp [I]) {dp [I] = dp [j] + 1 ;} if (ans <dp [I]) ans = dp [I] ;}} cout <
                   

                   

                   

                   
POJ1631 is also a naked LIS. The complexity is the nlogn template, and n ^ 2 times out.
                   


                   


                   

                   
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                                 # Define ll long # define eps 1e-7 # define inf 0 xfffffff // const ll INF = 1ll <61; using namespace std; // vector
                                
                                  > G; // typedef pair
                                 
                                   P; // vector
                                  
                                    >:: Iterator iter; // map
                                   
                                     Mp; // map
                                    
                                      : Iterator p; // # define IN freopen ("c: \ Users \ linzuojun \ desktop \ input.txt", "r", stdin) // # define OUT freopen ("c: \ Users \ linzuojun \ desktop \ output.txt", "w", stdout) int dp [100000 + 5]; int num [100000 + 5]; void clear () {memset (dp, 0, sizeof (dp); memset (num, 0, sizeof (num ));} int main () {int t, n; cin> t; while (t --) {clear (); cin> n; for (int I = 1; I <= n; I ++) cin> num [I]; int ans = 0; for (int I = 1; I <= n; I ++) {int tmp = num [I]; int left = 1; int right = ans; while (left <= right) {int mid = (left + right)/2; if (dp [mid] <tmp) left = mid + 1; elseright = mid-1;} dp [left] = tmp; if (ans <left) ans ++ ;} cout <
                                     

                                     

                                     

                                     
POJ1887: the maximum number of sub-sequences that do not rise is enough. In this case, pay attention to the output format. There is a leading space, and then there is a blank line between the two cases. It does not remind PE to say WA only.
                                     


                                     

                                     
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                                                  #define ll long long#define eps 1e-7#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector
                                                  
                                                    > G;//typedef pair
                                                   
                                                     P;//vector
                                                    
                                                      > ::iterator iter;////map
                                                     
                                                      mp;//map
                                                      
                                                       ::iterator p;//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin) //#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout) int dp[100000 + 5];int num[100000 + 5];void clear() {memset(dp,0,sizeof(dp));/*memset(num,0,sizeof(num));*/}int main() {int a;int Case = 0;clear();while(scanf("%d",&a)) {clear();int n = 1;if(a == -1)break;num[n] = a;while(true) {scanf("%d",&a);if(a == -1)break;num[++n] = a;}for(int i=1;i<=n;i++)dp[i] = 1;int ans = 1;for(int i=n-1;i>=1;i--) {for(int j=i+1;j<=n;j++) {if(dp[j] + 1 > dp[i] && num[i] > num[j])dp[i] = dp[j] + 1;if(dp[i] > ans)ans = dp[i];}}printf("Test #%d:\n",++Case);printf(" maximum possible interceptions: %d\n\n",ans);}return EXIT_SUCCESS;}
                                                      
                                                     
                                                    
                                                   
                                                  
                                                 
                                                
                                               
                                              
                                            
                                           
                                          
                                         
                                        
                                       
                                      
                                     

                                     

                                     

                                     
POJ1609, the longest sub-sequence is not dropped, but there are two restrictions. First, the sorting satisfies one of them, and then carry out the LIS Algorithm on this basis. Note that the non-equal sign must be changed, there are also the most questions, but the sequence is not given, so you need to sort first
                                     


                                     


                                     

                                     
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                                                  #define ll long long#define eps 1e-7#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector
                                                  
                                                    > G;//typedef pair
                                                   
                                                     P;//vector
                                                    
                                                      > ::iterator iter;////map
                                                     
                                                      mp;//map
                                                      
                                                       ::iterator p;//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin) //#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout) int dp[100000 + 5];struct Node {int x,y;}node[100000 + 5];void clear() {memset(dp,0,sizeof(dp));memset(node,0,sizeof(node));}bool cmp(Node x,Node y) {if(x.x == y.x)return x.y < y.y;return x.x 
                                                       
                                                         dp[i] && node[i].x >= node[j].x && node[i].y >= node[j].y)dp[i] = dp[j] + 1;if(dp[i] > ans)ans = dp[i];}}printf("%d\n",ans);}return EXIT_SUCCESS;}
                                                       
                                                      
                                                     
                                                    
                                                   
                                                  
                                                 
                                                
                                               
                                              
                                            
                                           
                                          
                                         
                                        
                                       
                                      
                                     

                                     

                                     
Another way is to do this, because the range is relatively small, so you can do it directly by using dp.
                                     


                                     

                                     
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                                                 #include
                                                 
                                                  #define ll long long#define eps 1e-7#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector
                                                  
                                                    > G;//typedef pair
                                                   
                                                     P;//vector
                                                    
                                                      > ::iterator iter;////map
                                                     
                                                      mp;//map
                                                      
                                                       ::iterator p;//#define IN freopen("c:\\Users\\linzuojun\\desktop\\input.txt", "r", stdin) //#define OUT freopen("c:\\Users\\linzuojun\\desktop\\output.txt", "w", stdout) int num[112][112],dp[112][112];void clear() {memset(num,0,sizeof(num));memset(dp,0,sizeof(dp));}int main() {int n;while(scanf("%d",&n),n) {clear();int x,y;for(int i=1;i<=n;i++) {scanf("%d %d",&x,&y); num[x][y]++;}for(int i=1;i<=100;i++) for(int j=1;j<=100;j++) dp[i][j] = max(dp[i-1][j],dp[i][j-1]) + num[i][j];printf("%d\n",dp[100][100]);}puts("*");return EXIT_SUCCESS;}