UVa 12299 线段树 单点更新 RMQ with Shifts

因为shift操作中的数不多，所以直接用单点更新模拟一下就好了。

太久不写线段树，手好生啊，不是这错一下就是那错一下。

PS：输入写的我有点蛋疼，不知道谁有没有更好的写法。

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 using namespace std;
  7 
  8 const int maxn = 100000 + 10;
  9 const int maxnode = (maxn << 2);
 10 const int INF = 0x3f3f3f3f;
 11 
 12 void scan(int& x)
 13 {
 14     x = 0;
 15     char c = ' ';
 16     while(c < '0' || c > '9') c = getchar();
 17     while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
 18 }
 19 
 20 int qL, qR;
 21 int p, v;
 22 int n, Q;
 23 int minv[maxnode];
 24 
 25 int a[maxn];
 26 
 27 void build(int o, int L, int R)
 28 {
 29     if(L == R) { scan(a[L]); minv[o] = a[L]; return ; }
 30     int M = (L + R) / 2;
 31     build(o<<1, L, M);
 32     build(o<<1|1, M+1, R);
 33     minv[o] = min(minv[o<<1], minv[o<<1|1]);
 34 }
 35 
 36 void update(int o, int L, int R)
 37 {
 38     //if(v >= minv[o]) return ;
 39     if(L == R) { minv[o] = v; return ; }
 40 
 41     int M = (L + R) / 2;
 42     if(p <= M) update(o<<1, L, M);
 43     else update(o<<1|1, M+1, R);
 44     minv[o] = min(minv[o<<1], minv[o<<1|1]);
 45 }
 46 
 47 int query(int o, int L, int R)
 48 {
 49     if(qL <= L && R <= qR) return minv[o];
 50     int ans = INF;
 51     int M = (L + R) / 2;
 52     if(qL <= M) ans = min(ans, query(o<<1, L, M));
 53     if(qR > M) ans = min(ans, query(o<<1|1, M+1, R));
 54     return ans;
 55 }
 56 
 57 char op[50];
 58 
 59 int main()
 60 {
 61     while(scanf("%d%d", &n, &Q) == 2 && n)
 62     {
 63         build(1, 1, n);
 64         while(Q--)
 65         {
 66             scanf("%s", op);
 67             int l = strlen(op);
 68             if(op[0] == 'q')
 69             {
 70                 int i = 6;
 71                 qL = 0;
 72                 while(i < l && op[i] >= '0' && op[i] <= '9') { qL = qL * 10 + op[i] - '0'; i++; }
 73                 while(i < l && (op[i] < '0' || op[i] > '9')) i++;
 74                 qR = 0;
 75                 while(i < l && op[i] >= '0' && op[i] <= '9') { qR = qR * 10 + op[i] - '0'; i++; }
 76                 printf("%d\n", query(1, 1, n));
 77             }
 78             else
 79             {
 80                 vector<int> hehe;
 81 
 82                 for(int i = 6; i < l;)
 83                 {
 84                     while(i < l && (op[i] < '0' || op[i] > '9')) i++;
 85                     if(i >= l) break;
 86                     int x = 0;
 87                     while(i < l && op[i] >= '0' && op[i] <= '9') { x = x * 10 + op[i] - '0'; i++; }
 88                     hehe.push_back(x);
 89                 }
 90 
 91                 int sz = hehe.size();
 92                 for(int i = 0; i < sz - 1; i++)
 93                 {
 94                     v = a[hehe[i + 1]];
 95                     p = hehe[i];
 96                     update(1, 1, n);
 97                 }
 98                 v = a[hehe[0]];
 99                 p = hehe[sz - 1];
100                 update(1, 1, n);
101 
102                 int t = a[hehe[0]];
103                 for(int i = 0; i < sz - 1; i++) a[hehe[i]] = a[hehe[i+1]];
104                 a[hehe[sz - 1]] = t;
105             }
106         }
107     }
108 
109     return 0;
110 }
代码君

 

posted @ 2015-08-05 11:29  AOQNRMGYXLMV  阅读(306)  评论(0)  编辑  收藏  举报