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ZOJ 3512 Financial Fraud (左偏树)
题意:给定一个序列,求另一个不递减序列,使得Abs(bi - ai) 和最小。
析:首先是在每个相同的区间中,中位数是最优的,然后由于要合并,和维护中位数,所以我们选用左偏树来维护,当然也可以用划分树来做。
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> #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 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 = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 50000 + 10; const int mod = 1e6 + 10; 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 int Abs(int x){ return x > 0 ? x : -x; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Node{ int key, l, r, d, fa; }; Node tr[maxn]; int iroot(int i){ if(i == -1) return -1; while(tr[i].fa != -1) i = tr[i].fa; return i; } int Merge(int rx, int ry){ if(rx == -1) return ry; if(ry == -1) return rx; if(tr[rx].key < tr[ry].key) swap(rx, ry); int r = Merge(tr[rx].r, ry); tr[rx].r = r; tr[r].fa = rx; if(tr[tr[rx].l].d < tr[r].d) swap(tr[rx].l, tr[rx].r); if(tr[rx].r == -1) tr[rx].d = 0; else tr[rx].d = tr[tr[rx].r].d + 1; return rx; } int del(int i){ if(i == -1) return -1; int l = tr[i].l, r = tr[i].r, y = tr[i].fa, x; tr[i].l = tr[i].r = tr[i].fa = -1; tr[x = Merge(l, r)].fa = y; if(y != -1 && tr[y].l == i) tr[y].l = x; else if(y != -1 && tr[y].r == i) tr[y].r = x; for( ; y != -1; x = y, y = tr[y].fa){ if(tr[tr[y].l].d < tr[tr[y].r].d) swap(tr[y].l, tr[y].r); if(tr[y].d == tr[tr[y].r].d + 1) break; tr[y].d = tr[tr[y].r].d + 1; } if(x != -1) return iroot(x); return iroot(y); } int top(int i){ return tr[i].key; } int pop(int &i){ Node out = tr[i]; int l = tr[i].l, r = tr[i].r; tr[i].l = tr[i].r = tr[i].fa = -1; tr[l].fa = tr[r].fa = -1; i = Merge(l, r); return out.key; } int a[maxn]; void init(){ for(int i = 0; i < n; ++i){ scanf("%d", a+i); tr[i].key = a[i]; tr[i].l = tr[i].r = tr[i].fa = -1; tr[i].d = 0; } } int tree[maxn], sz[maxn], cnt[maxn]; void solve(){ int m = -1; for(int i = 0; i < n; ++i){ tree[++m] = i; sz[m] = cnt[m] = 1; while(m > 0 && top(tree[m-1]) >= top(tree[m])){ tree[m-1] = Merge(tree[m-1], tree[m]); sz[m-1] += sz[m]; cnt[m-1] += cnt[m]; --m; while(cnt[m] > (sz[m]+1) / 2){ pop(tree[m]); --cnt[m]; } } } LL ans = 0; int k = 0; for(int i = 0; i <= m; ++i){ int t = top(tree[i]); for(int j = 0; j < sz[i]; ++j, ++k) ans += Abs(t - a[k]); } printf("%lld\n", ans); } int main(){ while(scanf("%d", &n) == 1 && n){ init(); solve(); } return 0; }
ZOJ 3512 Financial Fraud (左偏树)
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