POJ_3067 Japan[ 逆序数 树状数组 or 归并排序）

传送门：POJ_3067

题目：n,m,k;左右两列数，数的范围分别1-n，1-m，然给k个连线。

Sample Input

13 4 41 42 33 23 1

Sample Output

Test case 1: 5

思路：逆序数

代码：

//树状数组版

//树状数组#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>#include<string>#include<map>#include<vector>#include<queue>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define INF 0x3f3f3f3f#define PI acos(-1.0)#define mem(a,b) memset(a,b,sizeof a)#define read_ freopen("i.txt","r",stdin)using namespace std;typedef long long LL;const int N=1010;int n,m,k;int c[N];struct node{    int a,b;}p[1000010];int cmp(node a,node b){    if(a.a==b.a)        return a.b<b.b;    return a.a<b.a;}int lowbit(int x){    return x&(-x);}void add(int x,int v){    while(x<=m)    {        c[x]+=v;        x+=lowbit(x);    }}LL sum(int x){    LL ret=0;    while(x>0)    {        ret=ret+c[x];        x-=lowbit(x);    }    return ret;}int main(){    int T;    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        mem(c,0);        scanf("%d%d%d",&n,&m,&k);        for(int i=0;i<k;i++)            scanf("%d%d",&p[i].a,&p[i].b);        sort(p,p+k,cmp);        LL ans=0;        for(int i=0;i<k;i++){            add(p[i].b,1);            ans+=sum(m)-sum(p[i].b);        }        printf("Test case %d: %I64d\n",cas,ans);    }    return 0;}