1057. Stack (30)-PAT

1057. Stack (30)-中位数-BIT

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (N/2)-th smallest element if N is even, or ((N+1)/2)-th if N is odd.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<= 10⁵). Then N lines follow, each contains a command in one of the following 3 formats:

Push key
Pop
PeekMedian

where key is a positive integer no more than 10⁵.

Output Specification:

For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print "Invalid" instead.

Sample Input:

17PopPeekMedianPush 3PeekMedianPush 2PeekMedianPush 1PeekMedianPopPopPush 5Push 4PeekMedianPopPopPopPop

Sample Output:

InvalidInvalid322124453Invalid

推荐指数※※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1057

这道题如果不考虑时间限制，那是比较简单的，一开始就直接使用一个数组计数，使用二分法搜索，但都是超时。

这样只能用上统计累计频次比较好的Bianry Indexed Tree(http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees).

还有这里cin，cout还是不要用了，效率不如printf和scanf。

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;#define Maxval 100009int tree[Maxval+1];int stack[Maxval+1];void update(int idx,int val){while(idx<Maxval){tree[idx]+=val;idx+=(idx&-idx);}}int read(int idx){int sum=0;while(idx>0){sum+=tree[idx];idx-=(idx&-idx);}return sum;}int find_cumulate(int cumfre){int low=0,high=Maxval,mid;while(low<high-1){if((low+high)&1==1)mid=(low+high-1)/2;elsemid=(low+high)/2;int sum=read(mid);if(sum>=cumfre)high=mid;else  low=mid;}return high;}int main(){int n,i;char str[20];scanf("%d",&n);long tmp,top;memset(tree,0,Maxval*sizeof(int));memset(stack,0,Maxval*sizeof(int));top=0;for(i=0;i<n;i++){scanf("%str",str);if(str[1]=='o'){//POPif(top>0){tmp=stack[top--];printf("%d\n",tmp);update(tmp,-1);}elseprintf("Invalid\n");}else if(str[1]=='u'){//pushcin>>tmp;stack[++top]=tmp;update(tmp,1);}else if(str[1]=='e'){//mediaif(top>0){if(top&1==1)tmp=(top+1)/2;elsetmp=top/2;tmp=find_cumulate(tmp);printf("%d\n",tmp);}elseprintf("Invalid\n");}else{printf("Invalid\n");}}return 0;}

这是我超时的代码，比较的朴素

#include<iostream>#include<string.h>#include<string>#include<stack>#include<vector>#include<set>#include<algorithm>#define  N 100000#define  Hash_N 10000using namespace std;long find_median(long *arr,long first,long last,long val){long mid;while(first<last&&first!=0&&last!=Hash_N-1){mid=(first+last)/2;if(arr[mid]==val){return mid;}else if(arr[mid]>val)last=mid;elsefirst=mid;}return first;}int main(){long n,i,j;char str[20];long totalnum,num[N],hash_num[Hash_N];stack<long> st;cin>>n;totalnum=0;long tmp;memset(num,0,N*sizeof(int));memset(hash_num,0,Hash_N*sizeof(int));long fhash=N/Hash_N;for(i=0;i<n;i++){cin>>str;if(strcmp(str,"Pop")==0){//POPif(!st.empty()){tmp=st.top();cout<<tmp<<endl;st.pop();totalnum--;num[tmp]--;for(j=tmp/fhash;j<Hash_N;j++)hash_num[j]--;}elsecout<<"Invalid"<<endl;}else if(strcmp(str,"Push")==0){//pushcin>>tmp;st.push(tmp);totalnum++;num[tmp]++;hash_num[tmp/fhash]++;}else{//mediaif(!st.empty()){if(totalnum&1==1)tmp=(totalnum+1)/2;elsetmp=totalnum/2;long tmp_sum=0,k;k=find_median(hash_num,0,Hash_N,tmp);if(k>0)tmp_sum=hash_num[k-1];elsetmp_sum=0;for(j=k*fhash;j<N;j++){tmp_sum+=num[j];if(tmp_sum>=tmp){cout<<j<<endl;break;}}}elsecout<<"Invalid"<<endl;}}return 0;}