bzoj4430
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bit+容斥原理
我不会cdq分治只能用这个做法
考虑什么情况下不满足,至少有一个顺序不对就不行了,那么不满足的总有两对属性形成逆序对,那么我们用总方案数*2=n*(n-1)减去不符合的*2再/2就是答案
似乎进rank前200了
#include<bits/stdc++.h> using namespace std; const int N = 2e5 + 5; namespace IO { const int Maxlen = N * 50; char buf[Maxlen], *C = buf; int Len; inline void read_in() { Len = fread(C, 1, Maxlen, stdin); buf[Len] = ‘\0‘; } inline void fread(int &x) { x = 0; int f = 1; while (*C < ‘0‘ || ‘9‘ < *C) { if(*C == ‘-‘) f = -1; ++C; } while (‘0‘ <= *C && *C <= ‘9‘) x = (x << 1) + (x << 3) + *C - ‘0‘, ++C; x *= f; } inline void fread(long long &x) { x = 0; long long f = 1; while (*C < ‘0‘ || ‘9‘ < *C) { if(*C == ‘-‘) f = -1; ++C; } while (‘0‘ <= *C && *C <= ‘9‘) x = (x << 1) + (x << 3) + *C - ‘0‘, ++C; x *= f; } inline void read(int &x) { x = 0; int f = 1; char c = getchar(); while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) f = -1; c = getchar(); } while(c >= ‘0‘ && c <= ‘9‘) { x = (x << 1) + (x << 3) + c - ‘0‘; c = getchar(); } x *= f; } inline void read(long long &x) { x = 0; long long f = 1; char c = getchar(); while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) f = -1; c = getchar(); } while(c >= ‘0‘ && c <= ‘9‘) { x = (x << 1ll) + (x << 3ll) + c - ‘0‘; c = getchar(); } x *= f; } } using namespace IO; int n; int t[N], a[N], b[N], c[N], pos[N]; void update(int x, int d) { for(; x <= n; x += x & -x) t[x] += d; } int query(int x) { int ret = 0; for(; x; x -= x & -x) ret += t[x]; return ret; } long long solve(int *a, int *b) { long long ret = 0; for(int i = 1; i <= n; ++i) { pos[a[i]] = i; t[i] = 0; } for(int i = n; i; --i) { ret += query(pos[b[i]]); update(pos[b[i]], 1); } return ret; } int main() { read_in(); fread(n); for(int i = 1; i <= n; ++i) fread(a[i]); for(int i = 1; i <= n; ++i) fread(b[i]); for(int i = 1; i <= n; ++i) fread(c[i]); printf("%lld\n", ((long long)n * (long long)(n - 1) - solve(a, b) - solve(b, c) - solve(c, a)) >> 1); return 0; }
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