标签：
Joint StacksCrawling in process... Crawling failed Time Limit:4000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
Input
Output
Sample Input
Sample Output
Hint
Description
Input
, indicating the number of operations. The next N lines, each contain an instruction "push", "pop" or "merge". The elements of stacks are 32-bit integers. Both A and B are empty initially, and it is guaranteed that "pop" operation would not be performed to an empty stack. N = 0 indicates the end of input.Output
Sample Input
4 push A 1 push A 2 pop A pop A 9 push A 0 push A 1 push B 3 pop A push A 2 merge A B pop A pop A pop A 9 push A 0 push A 1 push B 3 pop A push A 2 merge B A pop B pop B pop B 0
Sample Output
Case #1: 2 1 Case #2: 1 2 3 0 Case #3: 1 2 3 0
/*今晚心事繁多，被一个E题卡了很久，心不在焉吧*/
#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> #include <string> #include <queue> #define N 50010 using namespace std; struct node { int num; int id; bool operator < (const node &other) const { return id<other.id; } }; priority_queue<node>a; priority_queue<node>b; priority_queue<node>c; /* 这里一定要用三个优先队列，因为当重复进行merge B A merge A B操作的时候一定会超时的（数据量太大了） 用三个优先队列的话，就能完美解决这个问题了，只要是合并的就合并到c中不管怎么进行操作，都不会再排序了 */ int n; char str[10];//命令 int main() { //freopen("in.txt","r",stdin); int p=1; while(~scanf("%d",&n)!=EOF&&n) { while(!a.empty()) a.pop(); while(!b.empty()) b.pop(); while(!c.empty()) c.pop(); printf("Case #%d:\n",p++); //memset(a,0,sizeof a); //memset(b,0,sizeof b); for(int i=0;i<n;i++) { scanf("%s",&str);//输入命令 //cout<<"str="<<str<<endl; char ch; int sh; if(strcmp(str,"push")==0) { //cout<<"第一个命令"<<endl; scanf("%*c%c %d",&ch,&sh); getchar(); //cout<<ch<<" "<<sh<<endl; if(ch==‘A‘) { node fr; fr.num=sh; fr.id=i; a.push(fr); } else if(ch==‘B‘) { node fr; fr.num=sh; fr.id=i; b.push(fr); } //cout<<str<<" "<<ch<<" "<<sh<<endl; } else if(strcmp(str,"pop")==0) { scanf("%*c%c",&ch); getchar(); if(ch==‘A‘) { if(a.empty()) { cout<<c.top().num<<endl; c.pop(); } else { cout<<a.top().num<<endl; a.pop(); } } else if(ch==‘B‘) { if(b.empty()) { cout<<c.top().num<<endl; c.pop(); } else { cout<<b.top().num<<endl; b.pop(); } } //cout<<str<<" "<<ch<<endl; } else if(strcmp(str,"merge")==0) { char s1,s2; scanf("%*c%c %c",&s1,&s2); getchar(); while(!a.empty()) { node fr=a.top(); a.pop(); c.push(fr); } while(!b.empty()) { node fr=b.top(); b.pop(); c.push(fr); } //cout<<str<<" "<<s1<<" "<<s2<<endl; } } } return 0; }
标签：
原文地址：http://www.cnblogs.com/wuwangchuxin0924/p/5762596.html
uDebug