hdu 5419 Victor and Toys 线段树成段更新

Victor and Toys

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 156    Accepted Submission(s): 54

Problem Description

Victor has n toys, numbered from 1 to n . The beauty of the i -th toy is wi .

Victor has a sense of math and he generates m intervals, the i -th interval is [li,ri] . He randomly picks 3 numbers i,j,k(1≤i<j<k≤m) , and selects all of the toys whose number are no less than max(li,lj,lk) and no larger than min(ri,rj,rk) . Now he wants to know the expected sum of beauty of the selected toys, can you help him?

 

Input

The first line of the input contains an integer T , denoting the number of test cases.

In every test case, there are two integers n and m in the first line, denoting the number of the toys and intervals.

The second line contains n integers, the i -th integer wi denotes that the beauty of the i -th toy.

Then there are m lines, the i -th line contains two integers li and ri .

1≤T≤10 .

1≤n,m≤50000 .

1≤wi≤5 .

1≤li≤ri≤n .

 

Output

Your program should print T lines : the i -th of these denotes the answer of the i -th case.

If the answer is an integer, just print a single interger, otherwise print an irreducible fraction like p/q .

 

Sample Input

   
   
   
   

    
    
    
    1
3 4
1 1 5
2 3
1 3
3 3
1 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5/4
   
   
   
   

 

Source

BestCoder Round #52 (div.2)

 

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题意，给出n个点，m个区间，先出三个区间，左值取最大，右值取最小，区间和的期望。

考虑每一个点，对结果的影响，如果，一个点包括了x个区间，那么，就有C(x,3)个w[i]在分子上，分母为C(m,3);所以最终的问题就是要求每个点，所点的区间数，真接用区间的开头加1，结尾减1，复杂度为o(n ),如果用线段树成段更新，复杂度为o(n * log(n);注意不要爆了long long,所以分子分母先都除上6.

第一种方法

#define N 300005
#define M 100005
#define maxn 205
#define SZ 26
#define MOD 1000000000000000007
#define lson (now<<1)
#define rson (now<<1|1)
int n,T,m,w[N],a,b,num[N],pri[N][2];
int main()
{
    while(S(T)!=EOF)
    {
        while(T--){
            S2(n,m);
            For(i,1,n+1){
                scan_d(a);
                w[i] = a;
                num[i] = 0;
                pri[i][0] =  pri[i][1] = 0;
            }
            FI(m){
                scan_d(a);scan_d(b);
                pri[a][0]++;
                pri[b][1]++;
            }
            int nn = 0;
            for(int i = 1;i<=n;i++){
                nn += pri[i][0];
                num[i] = nn;
                nn -= pri[i][1];
            }
            ll ans = 0,s2 = (ll)(m) * (ll)(m - 1) * (ll)(m -2) / 6;
            For(i,1,n+1){
                ll t = (ll) num[i];
                if(t >= 3){
                    ans += (ll)t * (ll)(t - 1) * (ll)(t - 2) * (ll)w[i] / 6;
                }
            }
            if(ans == 0){
                printf("%lld\n",ans);
                continue;
            }
            ll gc = gcd(ans,s2);
            ans /= gc;s2 /= gc;
            if(s2 == 1)
                printf("%lld\n",ans);
            else
                printf("%lld/%lld\n",ans,s2);
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

第二种方法（线段树）

#define N 300005
#define M 100005
#define maxn 205
#define SZ 26
#define MOD 1000000000000000007
#define lson (now<<1)
#define rson (now<<1|1)
int n,T,m,w[N],a,b;
struct node{
    int sum;
};
node tree[N*4];
void pushDown(int now){
    if(tree[now].sum > 0){
        tree[lson].sum  += tree[now].sum;
        tree[rson].sum += tree[now].sum;
        tree[now].sum = 0;
    }
}
void buildTree(int l,int r,int now){
    tree[now].sum = 0;
    if(l >= r){
        return ;
    }
    int mid = (l+r)>>1;
    buildTree(l,mid,lson);
    buildTree(mid+1,r,rson);
}
void updateTree(int l,int r,int now,int s,int e,int c){
    if(s <= l && e>= r){
        tree[now].sum += c;
        return ;
    }
    pushDown(now);
    int mid = (l+r)>>1;
    if(s <= mid) updateTree(l,mid,lson,s,e,c);
    if(e > mid) updateTree(mid+1,r,rson,s,e,c);
}
int queryTree(int l,int r,int now,int s,int e){
    if(s <= l && e>= r){
        return tree[now].sum;
    }
    pushDown(now);
    int mid = (l+r)>>1;
    if(s <= mid) return queryTree(l,mid,lson,s,e);
    if(e > mid) return queryTree(mid+1,r,rson,s,e);
    return 0;
}
int main()
{
    while(S(T)!=EOF)
    {
        while(T--){
            S2(n,m);
            buildTree(1,n,1);
            For(i,1,n+1){
                scan_d(a);
                w[i] = a;
            }
            FI(m){
                scan_d(a);scan_d(b);
                updateTree(1,n,1,a,b,1);
            }
            ll ans = 0,s2 = (ll)(m) * (ll)(m - 1) * (ll)(m -2) / 6;
            For(i,1,n+1){
                ll t = (ll) queryTree(1,n,1,i,i);
                if(t >= 3){
                    ans += (ll)t * (ll)(t - 1) * (ll)(t - 2) * (ll)w[i] / 6;
                }
            }
            if(ans == 0){
                printf("%lld\n",ans);
                continue;
            }
            ll gc = gcd(ans,s2);
            ans /= gc;s2 /= gc;
            if(s2 == 1)
                printf("%lld\n",ans);
            else
                printf("%lld/%lld\n",ans,s2);
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}