HDU 5792 World is Exploding

假设，只求一对pair（A,B)，就是一道简单的树状数组水题。只需要求出每个位置它右边有多少比它大的数，求一个前缀和就是答案。

但两对其实也一样，各求出两对的前缀和，相乘，然后再把a==c，a==d，b==c，b==d的不符情况减去。

先用树状数组预处理出以下四个数组：

lg[i]，左边比a[i]大的；

ls[i]，左边比a[i]小的；

rg[i]，右边比a[i]大的；

rs[i]，右边比a[i]小的；

怎么求不合要求的情况，可以这样考虑

假设a==c，A[b]在a的右边且A[b]>A[a]，即rg[a]，A[d]也在c的右边且A[d]<A[c]，即rs[c]；所以a==c的情况就是 rs[i]*rg[i];

其它类似。

【注意】a[i]的值可达1e9，所以要离散化，离散化的时候注意有重复的数字。

附上两份代码，另一份差不多是标程改过来的，其中树状数组的写法我搞不懂，问了某巨之后说是那样写树状数组的下标就可以从0开始了。

【代码1】

/* ***********************************************Author        :angon************************************************ */#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <stack>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define mst(a,k)  memset(a,k,sizeof(a))#define REP(i,k,n) for(int i=k;i<n;i++)#define REPP(i,k,n) for(int i=k;i<=n;i++)#define scan(d) scanf("%d",&d)#define scanl(d) scanf("%I64d",&d)#define scann(n,m) scanf("%d%d",&n,&m)#define scannl(n,m) scanf("%i64d%I64d",&n,&m)#define mst(a,k)  memset(a,k,sizeof(a))#define LL long long#define N 50010#define mod 1000000007inline int read(){int s=0;char ch=getchar();for(; ch<'0'||ch>'9'; ch=getchar());for(; ch>='0'&&ch<='9'; ch=getchar())s=s*10+ch-'0';return s;}LL a[N],b[N],ls[N],lg[N],rs[N],rg[N];int n,m;int c[N];int lowbit(int x){    return x & (-x);}void modify (int x,int add) {   //在x位置加上add，要修改所有祖先    while(x<=m)    {        c[x]+=add;        x+=lowbit(x);    }}LL get_sum(int x)  //统计前x项的和{    int ret=0;    while(x>0)    {        ret+=c[x];        x-=lowbit(x);    }    return ret;}int main(){    //freopen("1012.in","r",stdin);    //freopen("out.txt","w",stdout);    while(~scan(n))    {        REPP(i,1,n) scanl(a[i]),b[i]=a[i];        sort(b+1,b+n+1);        m = unique(b+1,b+n+1)-(b+1);        REPP(i,1,n) a[i] = lower_bound(b+1,b+m+1,a[i])-b;        LL sum1=0,sum2=0;        mst(c,0);        REPP(i,1,n)        {            ls[i] = get_sum(a[i]-1);            lg[i] = get_sum(m) - get_sum(a[i]);           // lg[i] = i - get_sum(a[i]);            sum1+=ls[i];            sum2+=lg[i];            modify(a[i],1);        }        mst(c,0);        for(int i=n;i>=1;i--)        {            rs[i] = get_sum(a[i]-1);            //rg[i] = n - i - get_sum(a[i]) +1;            rg[i] = get_sum(m) - get_sum(a[i]);            modify(a[i],1);        }        LL ans = sum1*sum2;        REPP(i,1,n)        {            ans -= rg[i]*rs[i];            ans -= rg[i]*lg[i];            ans -= ls[i]*rs[i];            ans -= ls[i]*lg[i];        }        printf("%I64d\n",ans);    }    return 0;}

【代码2】

#include <bits/stdc++.h>#define LL long long#define ALL(v) (v).begin(),(v).end()#define showtime fprintf(stderr,"time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)const int N = 100000 + 5;int n,A[N],B[N],C[N];int ls[N],rs[N],lg[N],rg[N];int tot;void modify(int p,int dt) {  //树状数组下标从0开始    for (int i = p; i < tot; i += ~i & i + 1) C[i] += dt;}int query(int p) {    int ret = 0;    for (int i = p; i >= 0; i -= ~i & i + 1) ret += C[i];    return ret;}int main() {    //freopen("1012.in","r",stdin);    //freopen("out.txt","w",stdout);    while (scanf("%d",&n) == 1) {        for (int i = 0; i < n; ++ i) {            scanf("%d",A + i);            B[i]=A[i];        }        sort(B,B+n);        tot = unique(B,B+n)-B;        for (int i = 0; i < n; ++ i) {            A[i] = lower_bound(B,B+tot,A[i]) - B ;        }        mst(C,0);        for (int i = 0; i < n; ++ i) {            ls[i] = query(A[i] - 1);            lg[i] = i - query(A[i]);            modify(A[i],1);        }        mst(C,0);        for (int i = n - 1; i >= 0; -- i) {            rs[i] = query(A[i] - 1);            rg[i] = n - 1 - i - query(A[i]);            modify(A[i],1);        }        LL result = 0,allg = accumulate(rs,rs + n,0ll);        for (int i = 0; i < n; ++ i) {            result += rg[i] * 1ll * (allg - lg[i] - rs[i]) - (lg[i] + rs[i]) * 1ll * ls[i];        }        printf("%I64d\n",result);    }    return 0;}