HDU-5792-World is Exploding(树状数组+离散化)
2016-08-03 11:03
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链接:http://acm.hdu.edu.cn/showproblem.php?pid=5792
题意:
给出一个长度为n的整数序列,求有多少个四元组{a,b,c,d}。
满足1<=a<b<=n,1<=c<d<=n ; a,b,c,d两两不相等,且 Va<Vb && Vc>Vd
题解:
对于a[i],预先处理出:
ll[i]: [0,i-1]小于a[i]的个数
lh[i]: [0,i-1]大于a[i]的个数
rl[i]: [i+1,n-1]小于a[i]的个数
rh[i]: [i+1,n-1]大于a[i]的个数
显然如果不考虑a,b,c,d两两不相等的情况时,答案ans=sigima(ll[i]+lh[i]);
然后减掉两两相等的情况就得到了终解。
![](https://img-blog.csdn.net/20160803105615119)
#include <bits/stdc++.h>
using namespace std;
const int maxn = 50010;
const int mod = 1e9+7;
int a[maxn],b[maxn];
int ll[maxn],lh[maxn],rl[maxn],rh[maxn];
int sum[maxn];
void add(int x ,int len)
{
while(x<=len){sum[x]++;x+=(x&-x);}
}
int query(int x)
{
int ans=0;
while(x>0){ans+=sum[x];x-=(x&-x);}
return ans;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0; i<n; ++i)
scanf("%d",&a[i]),b[i]=a[i];
sort(b,b+n);
int len = unique(b,b+n)-b;
for(int i=0; i<n; ++i)
a[i]=lower_bound(b,b+len,a[i])-b+1;
long long ans1=0,ans2=0;
for(int i=0; i<n; ++i)
{
ll[i]=query(a[i]-1);
lh[i]=i-query(a[i]);
ans1+=ll[i];
ans2+=lh[i];
add(a[i],len);
}
memset(sum,0,sizeof(sum));
for(int i=n-1; i>=0; --i)
{
rl[i]=query(a[i]-1);
rh[i]=n-1-i-query(a[i]);
add(a[i],len);
}
memset(sum,0,sizeof(sum));
long long ans=1LL*ans1*ans2;
for(int i=0; i<n; ++i)
{
ans-=1LL*rl[i]*rh[i];//a==c==a[i]
ans-=1LL*ll[i]*lh[i];//b==d==a[i]
ans-=1LL*ll[i]*rl[i];//b==c==a[i]
ans-=1LL*lh[i]*rh[i];//a==d==a[i]
}
cout<<ans<<endl;
}
return 0;
}
/*
4
2 4 1 3
4
1 2 3 4
*/
题意:
给出一个长度为n的整数序列,求有多少个四元组{a,b,c,d}。
满足1<=a<b<=n,1<=c<d<=n ; a,b,c,d两两不相等,且 Va<Vb && Vc>Vd
题解:
对于a[i],预先处理出:
ll[i]: [0,i-1]小于a[i]的个数
lh[i]: [0,i-1]大于a[i]的个数
rl[i]: [i+1,n-1]小于a[i]的个数
rh[i]: [i+1,n-1]大于a[i]的个数
显然如果不考虑a,b,c,d两两不相等的情况时,答案ans=sigima(ll[i]+lh[i]);
然后减掉两两相等的情况就得到了终解。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 50010;
const int mod = 1e9+7;
int a[maxn],b[maxn];
int ll[maxn],lh[maxn],rl[maxn],rh[maxn];
int sum[maxn];
void add(int x ,int len)
{
while(x<=len){sum[x]++;x+=(x&-x);}
}
int query(int x)
{
int ans=0;
while(x>0){ans+=sum[x];x-=(x&-x);}
return ans;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0; i<n; ++i)
scanf("%d",&a[i]),b[i]=a[i];
sort(b,b+n);
int len = unique(b,b+n)-b;
for(int i=0; i<n; ++i)
a[i]=lower_bound(b,b+len,a[i])-b+1;
long long ans1=0,ans2=0;
for(int i=0; i<n; ++i)
{
ll[i]=query(a[i]-1);
lh[i]=i-query(a[i]);
ans1+=ll[i];
ans2+=lh[i];
add(a[i],len);
}
memset(sum,0,sizeof(sum));
for(int i=n-1; i>=0; --i)
{
rl[i]=query(a[i]-1);
rh[i]=n-1-i-query(a[i]);
add(a[i],len);
}
memset(sum,0,sizeof(sum));
long long ans=1LL*ans1*ans2;
for(int i=0; i<n; ++i)
{
ans-=1LL*rl[i]*rh[i];//a==c==a[i]
ans-=1LL*ll[i]*lh[i];//b==d==a[i]
ans-=1LL*ll[i]*rl[i];//b==c==a[i]
ans-=1LL*lh[i]*rh[i];//a==d==a[i]
}
cout<<ans<<endl;
}
return 0;
}
/*
4
2 4 1 3
4
1 2 3 4
*/
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