HDU 1540 区间合并线段树

题目大意：

就是给定一堆位置，进行删除还原，最后找到 t 位置上的最大连续位置

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <iostream>
  4 
  5 using namespace std;
  6 const int N = 50005;
  7 
  8 struct Node{
  9     int l , r , ml , mr , ma; //ml左最长，mr右最长，ma总最长
 10 }tree[N<<2];
 11 
 12 void build(int o , int l , int r)
 13 {
 14     tree[o].l = l;
 15     tree[o].r = r;
 16     tree[o].ml = r - l + 1;
 17     tree[o].mr = r - l + 1;
 18     tree[o].ma = r - l + 1;
 19     int m = (l + r) / 2;
 20     if(l == r) return ;
 21     build(o<<1 , l , m);
 22     build(o<<1|1 , m+1 , r);
 23 }
 24 
 25 void push_up(int o)
 26 {
 27     int ls = o<<1 , rs = o<<1|1;
 28     tree[o].ml = tree[ls].ml;
 29     tree[o].mr = tree[rs].mr;
 30 
 31     if(tree[ls].ml == (tree[ls].r - tree[ls].l + 1))
 32         tree[o].ml = tree[ls].ml + tree[rs].ml;
 33 
 34     if(tree[rs].mr == (tree[rs].r - tree[rs].l + 1))
 35         tree[o].mr = tree[ls].mr + tree[rs].mr;
 36 
 37     tree[o].ma = max(tree[ls].mr + tree[rs].ml , tree[rs].ma);
 38     tree[o].ma = max(tree[ls].ma , tree[o].ma);
 39 }
 40 
 41 void update(int o , int t , int v)
 42 {
 43     if(tree[o].l == tree[o].r){
 44         if(v == 1) tree[o].ma = tree[o].ml = tree[o].mr = 1;
 45         else tree[o].ma = tree[o].ml = tree[o].mr = 0;
 46         return ;
 47     }
 48     int ls = o<<1 , rs = o<<1|1;
 49     int m = (tree[o].l + tree[o].r) / 2;
 50 
 51     if(t <= m) update(ls , t , v);
 52     else update(rs , t , v);
 53     //要等下层更新完才能递归回去更新上层
 54     push_up(o);
 55 }
 56 
 57 int query(int o , int t)
 58 {
 59     int ls = o<<1 , rs = o<<1|1 , m = (tree[o].l + tree[o].r) / 2;
 60     if(tree[o].l == tree[o].r || tree[o].ma == 0 || tree[o].ma == tree[o].r - tree[o].l + 1)
 61         return tree[o].ma;
 62 
 63     if(t <= m){
 64         /*此时t属于左子树，那么存在t只属于左子树区间，和属于左右子树合并的区间
 65         ，弱属于合并的区间，那么要保证，t到左子树右端点形成的连续的个数要小于
 66         左子树最大右端连续和
 67         也就是 m - t + 1 <= tree[ls].mr (这里m也等于tree[ls].r)
 68         */
 69         if(t >= tree[ls].r - tree[ls].mr + 1)
 70             return query(ls , t) + query(rs , m+1);
 71         else return query(ls , t);
 72     }else{
 73         if(t <= tree[rs].l + tree[rs].ml - 1)
 74             return query(ls , m) + query(rs , t);
 75         else query(rs , t);
 76     }
 77 }
 78 
 79 int que[N] , top;
 80 
 81 int main()
 82 {
 83  //   freopen("a.in" , "r" , stdin);
 84     int n,m;
 85     char str[10];
 86     int x;
 87     while(scanf("%d%d",&n,&m)!=EOF){
 88         build(1,1,n);
 89         top=0;
 90         while(m--)
 91         {
 92             scanf("%s",str);
 93             if(str[0]=='D'){
 94                 scanf("%d",&x);
 95                 que[top++]=x;
 96                 update(1,x,0);
 97             }
 98             else if(str[0]=='Q'){
 99                 scanf("%d",&x);
100                 printf("%d\n",query(1,x));
101             }
102             else{
103                 if(x>0){
104                     x=que[--top];
105                     update(1,x,1);
106                 }
107             }
108         }
109     }
110     return 0;
111 }

 

 posted on 2014-12-28 19:31  Love风吟  阅读(123)  评论(0)    收藏  举报