Hdu 2158 Good idea questions

Ideas:

Two variables I j for each I meet the search is maintained with J for the length of the interval is the minimum of I meet the requirements of the interval so you don't have to run o^n time complexity

#include <stdio.h> #include <string.h> #include <iostream>using namespaceStd; int Num[100010],Mark[100010]; int Main () {  int N,M,Q,I,J;  while (~scanf("%d%d",&N,&M)  ) {  if( N==0&&M==0 )  break;  for (I=1;I<=N;I++)scanf("%d",&Num[I]);  while (M--) {scanf("%d",&Q);Memset(Mark,0 ,sizeof( Mark)); int A; int Cont=0;  for (I=1;I<=Q;I++) {scanf("%d",&A); if (Mark[A]==0)Cont++;Mark[A]=1; }  int Leap[100010];Memset(Leap,0 ,sizeof( Leap)); int Min; int Flash=0;  for (I=1;;I + +) {  if( Mark[Num[I]]==0 )  continue;  for (J=I;J<=N;J + +) {  if( Mark[Num[J ]]) {  if( Leap[Num[J]]==0)Flash++;Leap[Num[J]]++; }  if( Flash==Cont )  break; }  break; }Min=J-I+1;Leap[Num[I]]--; if (Leap[Num[I]]==0)Flash--;  for (I++;I<=N;I + +) {  if(! Mark[Num[I ]])  continue; if (Flash==Cont ) {  if( J-I+1<Min)Min=J-I+1;Leap[Num[I]]--; if (Leap[Num[I]]==0)Flash--; Continue ; } for ( J++;J<=N;J + +) {  if(! Mark[Num[J ]])  continue; if (Leap[Num[J]]==0)Flash++;Leap[Num[J]]++; if (Flash==Cont )  break; }  if( Flash==Cont&&J-I+1<Min)Min=J-I+1;Leap[Num[I]]--; if (Leap[Num[I]]==0)Flash--; }Printf("%d\n",Min); }}  return 0;}

Hdu 2158 Good idea questions