BZOJ 2002 Bounce 弹飞绵羊 (分块或动态树)
2017-09-13 17:10
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题意:。。。
析:把这 n 个数进行分块,每次维护一块,当修改时,也是只是修改一块,查询是是块数。
代码如下:
析:把这 n 个数进行分块,每次维护一块,当修改时,也是只是修改一块,查询是是块数。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(x,n) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 200000 + 100; const int mod = 1000; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r > 0 && r <= n && c > 0 && c <= m; } int l[maxn], pos[maxn], val[maxn], r[maxn]; int a[maxn], belong[maxn]; int solve(int x){ int ans = 0; while(pos[x]){ ans += val[x]; x = pos[x]; } return ans + pos[x]; } int main(){ scanf("%d", &n); for(int i = 0; i < n; ++i) scanf("%d", a+i); int block = sqrt(n); int len = block; if(n % len) ++block; for(int i = 0; i < block; ++i) l[i] = i * len, r[i] = l[i] + len - 1; r[block-1] = n-1; // do not loss for(int i = 0; i < n; ++i) belong[i] = i / len; for(int i = n-1; i >= 0; --i) if(i + a[i] > r[belong[i]]) val[i] = 1, pos[i] = i + a[i]; else val[i] = val[i+a[i]] + 1, pos[i] = pos[i+a[i]]; scanf("%d", &m); while(m--){ int op, u, v; scanf("%d %d", &op, &u); if(op == 1) printf("%d\n", solve(u)); else{ scanf("%d", &v); a[u] = v; for(int i = u; i >= l[belong[u]]; --i) if(i + a[i] > r[belong[i]]) val[i] = 1, pos[i] = i + a[i]; else val[i] = val[i+a[i]] + 1, pos[i] = pos[i+a[i]]; } } return 0; }
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