HDU 6521 Party（线段树）

Posted 2020-11-26 xwdzuishuai

　　　题目意思：有n个人，一开始相互不认识。他们要去参加party，每次参加的人是编号在区间[l,r]内的人。参加完一次party之后，这区间内的人就会相互认识。每次会有有多少对人会新认识。那么用f[i] = j 表示 j - i的人都互相认识了，初始化f[i] = i,那么每次更新l,r话只需要找到是否存在f[l - r] > l并进行更新。

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 typedef long long ll;
  6 
  7 const int N = 5e5 + 10;
  8 
  9 int n, m;
 10 int val[N << 2], num[N  << 2];
 11 
 12 int read()
 13 {
 14     int res = 0, f = 1;
 15     char c = getchar();
 16     while(c < ‘0‘ || c > ‘9‘)
 17     {
 18         if(c == -1)
 19         {
 20             f = -1;
 21         }
 22         c = getchar(); 
 23     }
 24     while(c >= ‘0‘ && c <= ‘9‘)
 25     {
 26         res = res * 10 + c - ‘0‘;
 27         c = getchar();
 28     }
 29     return f * res;
 30 }
 31 
 32 void pushup(int rt)
 33 {
 34     val[rt] = max(val[rt << 1], val[rt << 1 | 1]);
 35 }
 36 
 37 void build(int rt, int l, int r)
 38 {
 39     if(l == r)
 40     {
 41         val[rt] = num[rt] = l;
 42         //puts("!");
 43         return ;
 44     }
 45     
 46     int mid = l + r >> 1;
 47     build(rt << 1, l, mid);
 48     build(rt << 1 | 1, mid + 1, r);
 49     pushup(rt);
 50 } 
 51 
 52 ll query(int L, int R, int l, int r, int rt, int x)
 53 {
 54     if(x >= val[rt]) return 0;
 55     if(l == r)
 56     {
 57         ll re = num[rt] - x;
 58         val[rt] = num[rt] = x;
 59         return re;
 60     }
 61     if(L <= l && r <= R)
 62     {
 63         if(x >= val[rt]) return 0;
 64         else
 65         {
 66             ll re = 0;
 67             int mid = l + r >> 1;
 68             re += query(L, min(R, mid), l, mid, rt << 1, x);
 69             re += query(max(L, mid + 1), R, mid + 1, r, rt << 1 | 1, x);
 70 
 71             pushup(rt);
 72             return re;
 73         }
 74     }
 75     ll res = 0;
 76     int mid = l + r >> 1;
 77     if(L <= mid)
 78     res += query(L, min(R, mid), l, mid, rt << 1, x);
 79     if(R > mid)
 80     res += query(max(L, mid + 1), R, mid + 1, r, rt << 1 | 1, x);
 81     pushup(rt);
 82     return res;
 83 }
 84 
 85 
 86 int main()
 87 {
 88     while(scanf("%d%d", &n, &m) != EOF)
 89     {
 90         build(1, 1, n);
 91         
 92     //    cout << val[1] <<" ! " << endl;
 93         int x, y;
 94         while(m --)
 95         {
 96         x = read(), y = read();
 97         ll ans = query(x, y, 1, n, 1, x);
 98         printf("%lld
", ans);
 99         }
100     }
101 }

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