Kiki & Little Kiki 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 873 Accepted Submission(s): 283

Problem Description Kiki is considered as a smart girl in HDU, many boys fall in love with her! Now, kiki will finish her education, and leave school, what a pity! One day, zjt meets a girl, who is like kiki very much, in the campus, and calls her little kiki. Now, little kiki want to design a container, which has two kinds of operation, push operation, and pop operation.
Push M:
Push the integer M into the container.
Pop M:
Find the maximal integer, which is not bigger than M, in the container. And pop it from the container. Specially, for all pop operations, M[i] is no bigger than M[i+1].
Although she is as smart as kiki, she still can't solve this problem! zjt is so busy, he let you to help little kiki to solve the problem. Can you solve the problem?

Input The input contains one or more data sets. At first line of each input data set is an integer N (1<= N <= 100000) indicate the number of operations. Then N lines follows, each line contains a word (“Push” or “Pop”) and an integer M. The word “Push” means a push operation, while “Pop” means a pop operation. You may assume all the numbers in the input file will be in the range of 32-bit integer.
Output For each pop operation, you should print the integer satisfy the condition. If there is no integer to pop, please print “No Element!”. Please print a blank line after each data set.
Sample Input

9
Push 10
Push 20
Pop 2
Pop 10
Push 5
Push 2
Pop 10
Pop 11
Pop 19
3
Push 2
Push 5
Pop 2

Sample Output

No Element!
10
5
2
No Element!

2

Source HDU 8th Programming Contest Site（1）
Recommend lcy | We have carefully selected several similar problems for you: 2276 2279 2278 2274 2277

一开始用离线处理+离散化+线段树的做法，然后一直错，后来想到会有重复元素，所以不适合这么做；
百度到了一个multiset的东西，可以完美地解决这个题，真是神奇啊

先贴一下线段树的代码

#include 
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               using namespace std; const int N = 100010; const int inf = -0x7fffffff; char str[10]; __int64 xis[N]; int cnt, x; __int64 ans; struct node { __int64 val; int x; int l, r; }tree[N << 2]; struct ope { bool flag; __int64 val; }can[N]; int Binsearch(__int64 val) { int l = 1, r = cnt, mid; while (l <= r) { mid = (l + r) >> 1; if (xis[mid] == val) { break; } else if (xis[mid] > val) { r = mid - 1; } else { l = mid + 1; } } return mid; } void build(int p, int l, int r) { tree[p].l = l; tree[p].r = r; tree[p].val = inf; if (l == r) { tree[p].x = l; return; } int mid = (l + r) >> 1; build(p << 1, l, mid); build(p << 1 | 1, mid + 1, r); } void insert(int p, int pos, __int64 val) { if (tree[p].l == tree[p].r) { tree[p].val = val; return; } int mid = (tree[p].l + tree[p].r) >> 1; if (pos <= mid) { insert(p << 1, pos, val); } else { insert(p << 1 | 1, pos, val); } if (tree[p << 1].val > tree[p << 1 | 1].val) { tree[p].val = tree[p << 1].val; tree[p].x = tree[p << 1].x; } else { tree[p].val = tree[p << 1 | 1].val; tree[p].x = tree[p << 1 | 1].x; } } void query(int p, int l, int r) { if (tree[p].l >= l && r >= tree[p].r) { if (ans < tree[p].val) { ans = tree[p].val; x = tree[p].x; } return; } int mid = (tree[p].l + tree[p].r) >> 1; if (r <= mid) { query(p << 1, l, r); } else if (l > mid) { query(p << 1 | 1, l, r); } else { query(p << 1, l, mid); query(p << 1 | 1, mid + 1, r); } } void del(int p, int pos) { if (tree[p].l == tree[p].r) { tree[p].val = inf; return; } int mid = (tree[p].l + tree[p].r) >> 1; if (pos <= mid) { del(p << 1, pos); } else { del(p << 1 | 1, pos); } if (tree[p << 1].val > tree[p << 1 | 1].val) { tree[p].val = tree[p << 1].val; tree[p].x = tree[p << 1].x; } else { tree[p].val = tree[p << 1 | 1].val; tree[p].x = tree[p << 1 | 1].x; } } int main() { int n; while (~scanf("%d", &n)) { cnt = 0; __int64 val; for (int i = 0; i < n; ++i) { scanf("%s%I64d", str, &val); if (!strcmp(str, "Push")) { can[i].flag = 1; } else { can[i].flag = 0; } can[i].val = val; xis[++cnt] = val; } sort (xis + 1, xis + 1 + cnt); cnt = unique(xis + 1, xis + 1 + cnt) - xis - 1; build(1, 1, cnt); for (int i = 0; i < n; ++i) { int pos = Binsearch(can[i].val); if (can[i].flag) { insert(1, pos, can[i].val); } else { ans = inf; query(1, 1, pos); if (ans == inf) { printf("No Element!\n"); } else { printf("%I64d\n", ans); del(1, x); } } } printf("\n"); } return 0; }

AC代码

#include 
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               using namespace std; multiset
              
               st; multiset
               
                 :: iterator it; char str[10]; int main() { int n, m; while (~scanf("%d", &n)) { st.clear(); for (int i = 0; i < n; i++) { scanf("%s%d", str, &m); if (!strcmp(str, "Push")) { st.insert(m); } else { it = st.upper_bound(m); if (it == st.begin()) { printf("No Element!\n"); } else { it--; printf("%d\n", *it); st.erase(it); } } } printf("\n"); } return 0; }

hdu2275――Kiki & Little Kiki 1

Kiki & Little Kiki 1