NOIP2013 Match Line

The classic topic of reverse order. Exam time do not know how to write ugly, all WA, just take me to review the reverse order.

The match sequence is assigned a rank from small to large, and when the grade of a is at the same level as the corresponding B, the answer is the least, and as for why, I do not prove it. The hierarchy here is, in fact, discretization.

The rank of a from small to large, and then the b corresponding to the rank of a, the order appears in the rank sequence of B pairs, is the number of exchanges we require, because each exchange, can only make a group of reverse order to become orderly.

Note that the search for reverse order is to time out, only 70 points, so the tree-like array can be used.

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define MAXN    100005#define mod 99999997using namespace std;struct t{int num; int POS;} A[maxn],b[maxn];int tree[maxn];int c[maxn];int n;bool cmp1 (T x,t y) {return x.num < Y.num;} int lowbit (int x) {return x&-x;} void Update (int pos) {for (int i = pos; I <= n; i + = Lowbit (i)) tree[i]++;}    int getsum (int pos) {int res = 0;    for (int i = pos; I >= 1; I-= Lowbit (i)) res = (Res+tree[i])%mod; return res;}    int main () {scanf ("%d", &n);    for (int i = 1; I <= n; i++) {scanf ("%d", &a[i].num); a[i].pos = i;}    for (int i = 1; I <= n; i++) {scanf ("%d", &b[i].num); b[i].pos = i;}    Sort (A+1,A+N+1,CMP1);    Sort (B+1,B+N+1,CMP1);    for (int i = 1; I <= n; i++) C[a[i].pos] = B[i].pos;    int ans = 0;        for (int i = 1; I <= n; i++) {update (c[i]);    Ans = (ans + i-getsum (c[i]))%mod;    } printf ("%d\n", ans); Return 0;} 

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NOIP2013 Match Line