洛谷P2221 高速公路【线段树】

题目：https://www.luogu.org/problemnew/show/P2221

题意：有n个节点排成一条链，相邻节点之间有一条路。

C u v val表示从u到v的路径上的每条边权值都加val。

Q l r表示在l到r中等概率选择两个城市的路径长度的期望值。

思路：首先期望值的分子肯定是可以选择的方案数也就是$C^2_{r - l + 1}$

分子应该是所有可能的路径和。我们可以通过计算每一条边算了多少次得到。

对于第$i$条边，他的左端点有$(i - l + 1)$种可能，右端点有$(r - i + 1)$种可能。因此这$(i - l + 1)*(r - i + 1)$种路径都包含第$i$条边

所以分子可以表示为$\sum_{l}^{r}(i-l+1)*(r - i + 1) * a[i]$

把含$i$的和不含的都分离出来。可以变为$(r - l + 1-r*l)\sum a[i] + (r + l)\sum i *a[i] - \sum i^2*a[i]$

分别用线段树维护$\sum a[i], \sum i * a[i], \sum i^2 * a[i]$

小trick是$i$和$i^2$之和也可以保存在线段树节点中，维护起来比较方便。

 #include<cstdio>
 #include<cstdlib>
 #include<map>
 #include<set>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<cmath>
 #include<stack>
 #include<queue>
 #include<iostream>

 #define inf 0x3f3f3f3f
 using namespace std;
 typedef long long LL;
 typedef pair<int, int> pr;

 int n, m;
 const int maxn = 1e5 + ;
 struct node{
     LL i, a, ia, ii, iia, lazy;
 }tree[maxn * ];

 void build(int rt, int l, int r)
 {
     if(l == r){
         tree[rt].i = l;
         tree[rt].ii = 1ll * l * l;
         return;
     }
     int mid = (l + r) / ;
     build(rt << , l, mid);
     build(rt <<  | , mid + , r);
     tree[rt].i = tree[rt << ].i + tree[rt <<  | ].i;
     tree[rt].ii = tree[rt << ].ii + tree[rt <<  | ].ii;
 }

 void pushdown(int rt, int l, int r)
 {
     if(tree[rt].lazy){
         tree[rt << ].lazy += tree[rt].lazy;
         tree[rt <<  | ].lazy += tree[rt].lazy;
         int mid = (l + r) / ;
         tree[rt << ].a += 1ll * tree[rt].lazy * (mid - l + );
         tree[rt <<  | ].a += 1ll * tree[rt].lazy * (r - mid);
         tree[rt << ].ia += 1ll * tree[rt].lazy * tree[rt << ].i;
         tree[rt << |].ia += 1ll * tree[rt].lazy * tree[rt << |].i;
         tree[rt << ].iia += 1ll * tree[rt].lazy * tree[rt << ].ii;
         tree[rt << |].iia += 1ll * tree[rt].lazy * tree[rt << |].ii;
         tree[rt].lazy = ;
     }

 }

 void pushup(int rt)
 {
     tree[rt].a = tree[rt << ].a + tree[rt <<  |].a;
     tree[rt].ia = tree[rt << ].ia + tree[rt <<  | ].ia;
     tree[rt].iia = tree[rt << ].iia + tree[rt <<  | ].iia;
 }

 void update(int L, int R, int l, int r, int rt, int val)
 {
     if(L <= l && R >= r){
         tree[rt].a += 1ll * val * (r - l + );
         tree[rt].ia += 1ll * val * tree[rt].i;
         tree[rt].iia += 1ll * val * tree[rt].ii;
         tree[rt].lazy += val;
         return;
     }
     pushdown(rt, l, r);
     int mid = (l + r) / ;
     if(L <= mid)update(L, R, l, mid, rt << , val);
     if(R > mid)update(L, R, mid + , r, rt <<  | , val);
     pushup(rt);
 }

 LL sum1, sum2, sum3;
 void query(int L, int R, int l, int r, int rt)
 {
     if(L <= l && R >= r){
         sum1 += tree[rt].a;
         sum2 += tree[rt].ia;
         sum3 += tree[rt].iia;
         return;
     }
     pushdown(rt, l, r);
     int mid = (l + r) / ;
     if(L <= mid)query(L, R, l, mid, rt << );
     if(R > mid)query(L, R, mid + , r, rt <<  | );

 }

 LL gcd(LL a, LL b)
 {
     if(!b)return a;
     else return gcd(b, a % b);
 }
 int main()
 {
     scanf("%d%d", &n, &m);
     build(, , n - );
     for(int i = ; i < m; i++){
         string type;
         int l, r;
         cin>>type>>l>>r; r--;
         if(type[] == 'C'){
             int val;
             scanf("%d", &val);
             update(l, r, , n - , , val);
         }
         else{
             sum1 = sum2 = sum3 = ;
             query(l, r, , n - , );
             LL fac = (1ll * r - l +  - 1ll * r * l) * sum1 + (r + l) * sum2 - sum3;
             LL div = 1ll * (r - l + ) * (r - l + ) / ;
 //            cout<<sum1<<" "<<sum2<<" "<<sum3<<endl;
 //            cout<<fac<<" "<<div<<endl;
             LL g = gcd(fac, div);
             fac /= g;
             div /= g;
             printf("%lld/%lld\n", fac, div);
         }
     }
     return ;
 }