luogu P1966 火柴排队 (逆序对)

Posted 2021-01-09 tpgzy

luogu P1966 火柴排队

题目链接:https://www.luogu.org/problemnew/show/P1966
显然贪心的想,排名一样的数相减是最优的.
证明也很简单.
此处就不证明了.
然后交换的话就是求一个逆序对.
怎么样排序是一个关键.
(c)数组的下标是(a)的排名,值是(b)的值.
这样求逆序对的时候,就是排名为(i)的(a)数组,会对应上相应排名的(b)数组的上.
这也算是一个小技巧吧.

#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;

const int maxN = 100000 + 7;
const int mod = 99999997;

int n;
int f[maxN];
struct Node {
    int id,w;
}a[maxN],b[maxN];
int c[maxN];
int ans;

inline int read() {
    int x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}
    while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
    return x * f;
}

bool cmp(Node a,Node b) {
    return a.w < b.w;
}

inline int lowbit(int x) {return x & -x;}

void modify(int pos,int val) {
    while(pos <= n) f[pos] += val,pos += lowbit(pos);
} 

int query(int pos) {
    int sum = 0;
    while(pos) sum += f[pos],pos -= lowbit(pos); 
    return sum;
}

void re_pair1() {
    for(int i = 1;i <= n;++ i) {
        modify(c[i],1);
        ans += query(n) - query(c[i]);
        ans %= mod;
    }
    printf("%d", ans);
    return ;
}

int main() {
    n = read();
    for(int i = 1;i <= n;++ i) 
        a[i].id = i,a[i].w = read();
    for(int i = 1;i <= n;++ i) 
        b[i].id = i,b[i].w = read();
    sort(b + 1,b + n + 1,cmp);
    sort(a + 1,a + n + 1,cmp);
    for(int i = 1;i <= n;++ i)
        c[a[i].id] = b[i].id;
    re_pair1();
    return 0;
}