[20180824] noip模拟赛 Challenge-优快云博客

线段树？分块大法好

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 //#define ll int
  5 const int N = 2e5 + 10;
  6 
  7 int n, m, a[N];
  8 
  9 int L[N], R[N], bel[N], len, cnt;
 10 
 11 int p[500];
 12 
 13 ll sum[450], tag[450], vl[N], dat[N][450];
 14 
 15 void sol(int pos, int val) {
 16     
 17 }
 18 
 19 inline int query() {
 20     for(int i = 1 ; i <= cnt ; ++ i) {
 21         ll val = -tag[i];
 22         if(dat[i][1] <= val && val <= dat[i][p[i]]) {
 23             ll fn = *lower_bound(dat[i] + 1, dat[i] + 1 + p[i], val);
 24             if(fn == val) {
 25                 for(int j = L[i] ; j <= R[i] ; ++ j) {
 26                     if(vl[j] == val) {
 27                         return j;
 28                     }
 29                 }
 30             }
 31         }
 32     }
 33     return -1;
 34 }
 35 
 36 struct FastIO {
 37     static const int S = 1e7;
 38     int wpos;
 39     char wbuf[S];
 40     FastIO() : wpos(0) {}
 41     inline int xchar() {
 42         static char buf[S];
 43         static int len = 0, pos = 0;
 44         if (pos == len)
 45             pos = 0, len = fread(buf, 1, S, stdin);
 46         if (pos == len) exit(0);
 47         return buf[pos++];
 48     }
 49     inline int xuint() {
 50         int c = xchar(), x = 0;
 51         while (c <= 32) c = xchar();
 52         for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
 53         return x;
 54     }
 55     ~FastIO()
 56     {
 57         if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
 58     }
 59 } io;
 60 
 61 template<typename T> void read(T &x) {
 62     char c = x = 0;
 63     while(!isdigit(c)) c = getchar();
 64     while(isdigit(c)) x = x * 10 + c - '0', c = getchar();
 65 }
 66 
 67 int main() {
 68     n = io.xuint(), m = io.xuint();
 69     len = min(n, int(sqrt(n)) + 1);
 70     cnt = (n - 1) / len + 1;
 71     ll s = 0;
 72     for(int i = 1 ; i <= n ; ++ i) {
 73         int id = (i - 1) / len + 1;
 74         if(!L[id]) L[id] = i; R[id] = i;
 75         bel[i] = id, a[i] = io.xuint();
 76         dat[id][++ p[id]] = s - a[i];
 77         vl[i] = s - a[i];
 78         sum[id] += a[i];
 79         s += a[i];
 80     }
 81     for(int i = 1 ; i <= cnt ; ++ i) {
 82         sort(dat[i] + 1, dat[i] + 1 + p[i]);
 83         sum[i] += sum[i - 1];
 84     }
 85     for(int i = 1, pos, val ; i <= m ; ++ i) {
 86         pos = io.xuint(), val = io.xuint();
 87         
 88         val = val - a[pos];
 89         if(val) {
 90             a[pos] += val;
 91             int id = bel[pos];
 92             p[id] = 0;
 93             ll s = sum[id - 1];
 94             
 95             for(int i = L[id] ; i <= R[id]; ++ i) {    
 96                 dat[id][++ p[id]] = s - a[i];
 97                 vl[i] = s - a[i];
 98                 s += a[i];
 99             }
100             
101             sort(dat[id] + 1, dat[id] + 1 + p[id]);
102             
103             tag[id] = 0;
104             
105             sum[id] += val;
106             for(int i = id + 1 ; i <= cnt ; ++ i) {
107                 sum[i] += val;
108                 tag[i] += val;
109             }
110         }
111         printf("%d\n", query());
112     }
113 }