题目传送门

 /*
这题可以用stl的mutiset容器方便求解,我对这东西不熟悉,TLE了几次,最后用读入外挂水过。
题解有O(n)的做法,还以为我是侥幸过的,后来才知道iterator it写在循环内才超时了,囧!
*/
/************************************************
Author :Running_Time
Created Time :2015-8-4 12:10:11
File Name :G.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + ;
multiset<int>::iterator it; inline int read(void)
{
int x = , f = ; char ch = getchar ();
while (ch < '' || ch > '') {if (ch == '-') f = -; ch = getchar ();}
while (ch >= '' && ch <= '') {x = x * + ch - ''; ch = getchar ();}
return x * f;
} int main(void) { //HDOJ 5349 MZL's simple problem
int n;
while (scanf ("%d", &n) == ) {
multiset<int> S;
for (int i=; i<=n; ++i) {
int op, x;
op = read ();
if (op == ) {
x = read ();
S.insert (x);
}
else if (op == ) {
if (S.empty ()) continue;
S.erase (S.begin ());
}
else if (op == ) {
if (S.empty ()) {
puts (""); continue;
}
it = S.end (); it--;
printf ("%d\n", *it);
}
}
} return ;
}

 /************************************************
* Author :Running_Time
* Created Time :2015-8-5 16:42:43
* File Name :G_2.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e5 + ;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + ; inline int read(void)
{
int x = , f = ; char ch = getchar ();
while (ch < '' || ch > '') {if (ch == '-') f = -; ch = getchar ();}
while (ch >= '' && ch <= '') {x = x * + ch - ''; ch = getchar ();}
return x * f;
} int main(void) {
int n; n = read ();
int sz = , mx = -2e9;
for (int i=; i<=n; ++i) {
int op, x; op = read ();
if (op == ) {
x = read ();
sz++; mx = max (mx, x);
}
else if (op == ) {
sz = max (, sz - );
if (!sz) mx = -2e9;
}
else {
if (!sz) puts ("");
else printf ("%d\n", mx);
}
} return ;
}

标程做法

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