题目1522：包含min函数的栈

http://ac.jobdu.com/problem.php?pid=1522

我想到了用multiset的方法

#include<iostream>
#include<set>
#include<stack>
#include<stdio.h>
#include<queue>
#include<algorithm>

using namespace std;

struct data{
    int v;
    friend bool operator <(data a,data b){
        return a.v<b.v;
    }
    data(int a){
        v=a;
    }

};

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        int i;
        multiset <data>set1;
        stack<int>sta;

        char s[9];
        int temp;
        for(i=1;i<=n;i++){
            scanf("%s",s);
            if(s[0]=='s'){
                scanf("%d",&temp);
                sta.push(temp);
                set1.insert(temp);
            }
            else{
                if(!set1.empty()){
                    multiset <data>::iterator po;
                    for(po=set1.begin();po!=set1.end();po++){
                        if(po->v==sta.top())break;
                    }

                    set1.erase(po);
                    sta.pop();
                }
            }

            if(set1.begin()!=set1.end())
                printf("%d
",set1.begin()->v);//写成*set1.begin()编译可以过 但会RE
            else
                printf("NULL
");
        }
    }

    return 0;
}

 但是对SET里面的数据处理时不能直接删除数字，而是要删除数字所在的位置

还有一种比较简单的方法，就是不管如何，把较小的数字入队列

#include<stdio.h>

int sta[1009999];

int min(int a,int b){
    if(a>b)return b;
    else return a;
}
int main()
{
    int n;
    sta[0]=999999999;
    while(scanf("%d",&n)!=EOF){
        int i,add=0;
        char s[9];
        int temp;
        for(i=1;i<=n;i++){
            scanf("%s",s);
            if(s[0]=='s'){
                add++;
                scanf("%d",&temp);
                sta[add]=min(sta[add-1],temp);
                printf("%d
",sta[add]);
                
            }else{
                add--;
                if(sta[add]==999999999)
                    printf("NULL
");
                else
                    printf("%d
",sta[add]);
            }
        }
    }

    return 0;
}