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1 /*
2 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。
3 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！
4 */
5 /************************************************
6 Author :Running_Time
7 Created Time :2015-8-4 12:10:11
8 File Name :G.cpp
9 ************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 1e5 + 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36 multiset<int>::iterator it;
37
38 inline int read(void)
39 {
40 int x = 0, f = 1; char ch = getchar ();
41 while (ch < ‘0‘ || ch > ‘9‘) {if (ch == ‘-‘) f = -1; ch = getchar ();}
42 while (ch >= ‘0‘ && ch <= ‘9‘) {x = x * 10 + ch - ‘0‘; ch = getchar ();}
43 return x * f;
44 }
45
46 int main(void) { //HDOJ 5349 MZL‘s simple problem
47 int n;
48 while (scanf ("%d", &n) == 1) {
49 multiset<int> S;
50 for (int i=1; i<=n; ++i) {
51 int op, x;
52 op = read ();
53 if (op == 1) {
54 x = read ();
55 S.insert (x);
56 }
57 else if (op == 2) {
58 if (S.empty ()) continue;
59 S.erase (S.begin ());
60 }
61 else if (op == 3) {
62 if (S.empty ()) {
63 puts ("0"); continue;
64 }
65 it = S.end (); it--;
66 printf ("%d\n", *it);
67 }
68 }
69 }
70
71 return 0;
72 }
1 /************************************************ 2 * Author :Running_Time 3 * Created Time :2015-8-5 16:42:43 4 * File Name :G_2.cpp 5 ************************************************/ 6 7 #include <cstdio> 8 #include <algorithm> 9 #include <iostream> 10 #include <sstream> 11 #include <cstring> 12 #include <cmath> 13 #include <string> 14 #include <vector> 15 #include <queue> 16 #include <deque> 17 #include <stack> 18 #include <list> 19 #include <map> 20 #include <set> 21 #include <bitset> 22 #include <cstdlib> 23 #include <ctime> 24 using namespace std; 25 26 #define lson l, mid, rt << 1 27 #define rson mid + 1, r, rt << 1 | 1 28 typedef long long ll; 29 const int MAXN = 1e5 + 10; 30 const int INF = 0x3f3f3f3f; 31 const int MOD = 1e9 + 7; 32 33 inline int read(void) 34 { 35 int x = 0, f = 1; char ch = getchar (); 36 while (ch < ‘0‘ || ch > ‘9‘) {if (ch == ‘-‘) f = -1; ch = getchar ();} 37 while (ch >= ‘0‘ && ch <= ‘9‘) {x = x * 10 + ch - ‘0‘; ch = getchar ();} 38 return x * f; 39 } 40 41 int main(void) { 42 int n; n = read (); 43 int sz = 0, mx = -2e9; 44 for (int i=1; i<=n; ++i) { 45 int op, x; op = read (); 46 if (op == 1) { 47 x = read (); 48 sz++; mx = max (mx, x); 49 } 50 else if (op == 2) { 51 sz = max (0, sz - 1); 52 if (!sz) mx = -2e9; 53 } 54 else { 55 if (!sz) puts ("0"); 56 else printf ("%d\n", mx); 57 } 58 } 59 60 return 0; 61 }
mutiset HDOJ 5349 MZL's simple problem
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原文地址：http://www.cnblogs.com/Running-Time/p/4705213.html
