P2221 [HAOI2012]高速公路

思路

考虑每一条边的贡献，然后推式子
\[ \begin{align}&\sum_{i}V_i\times(R-i+1)\times(i-L+1)\\=&\sum_{i}V_i\left[(Ri-i^2+i)-(RL-iL+L)+(R-i+1)\right]\\=&\sum_{i}V_i\left[Ri-i^2+i-RL+Li-L+R-i+1\right]\\=&\sum_{i}Vi\left[(Ri+Li)-i^2-RL+(R-L+1)\right]\\=&\sum_{i}(Ri+Li)V_i-\sum_{i}i^2V_i-\sum_{i}RLV_i+\sum_{i}(R-L+1)V_i\\=&(R+L)\sum_{i}iVi-\sum_{i}i^2V_i-RL\sum_{i}V_i+(R-L+1)\sum_{i}V_i\\=&(R+L)\sum_iiV_i-\sum_ii^2V_i+(R-L+1-RL)\sum_{i}V_i\end{align} \]
然后用线段树维护就好了

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#define int long long
using namespace std;
const int MAXN = 100100;
namespace Seg1{//v_i
    int seg[MAXN<<2]={},tag[MAXN<<2]={};
    void pushup(int o){
        seg[o]=seg[o<<1]+seg[o<<1|1];
    }
    void build(int l,int r,int o,int *a){
        if(l==r){
            seg[o]=a[l];
            return;
        }
        int mid=(l+r)>>1;
        build(l,mid,o<<1,a);
        build(mid+1,r,o<<1|1,a);
        pushup(o);
    }
    void pushdown(int o,int ln,int rn){
        if(tag[o]){
            seg[o<<1]+=tag[o]*ln;
            seg[o<<1|1]+=tag[o]*rn;
            tag[o<<1]+=tag[o];
            tag[o<<1|1]+=tag[o];
            tag[o]=0;
        }
    }    
    void add(int L,int R,int l,int r,int o,int c){
        if(L<=l&&r<=R){
            seg[o]+=c*(r-l+1);
            tag[o]+=c;
            return;
        }
        int mid=(l+r)>>1;
        pushdown(o,mid-l+1,r-mid);
        if(L<=mid)
            add(L,R,l,mid,o<<1,c);
        if(R>mid)
            add(L,R,mid+1,r,o<<1|1,c);
        pushup(o);
    }
    int query(int L,int R,int l,int r,int o){
        if(L<=l&&r<=R){
            return seg[o];
        }
        int mid=(l+r)>>1,ans=0;
        pushdown(o,mid-l+1,r-mid);
        if(L<=mid)
            ans+=query(L,R,l,mid,o<<1);
        if(R>mid)
            ans+=query(L,R,mid+1,r,o<<1|1);
        return ans;
    }
};
namespace Seg2{//v_i*i
    int seg[MAXN<<2]={},tag[MAXN<<2]={};
    int sum(int l,int r){
        return (l+r)*(r-l+1)/2;
    }
    void pushup(int o){
        seg[o]=seg[o<<1]+seg[o<<1|1];
    }
    void build(int l,int r,int o,int *a){
        if(l==r){
            seg[o]=a[l]*l;
            return;
        }
        int mid=(l+r)>>1;
        build(l,mid,o<<1,a);
        build(mid+1,r,o<<1|1,a);
        pushup(o);
    }
    void pushdown(int o,int lx,int rx){
        if(tag[o]){
            int mid=(lx+rx)>>1;
            seg[o<<1]+=tag[o]*sum(lx,mid);
            seg[o<<1|1]+=tag[o]*sum(mid+1,rx);
            tag[o<<1]+=tag[o];
            tag[o<<1|1]+=tag[o];
            tag[o]=0;
        }
    }    
    void add(int L,int R,int l,int r,int o,int c){
        if(L<=l&&r<=R){
            seg[o]+=c*sum(l,r);
            tag[o]+=c;
            return;
        }
        int mid=(l+r)>>1;
        pushdown(o,l,r);
        if(L<=mid)
            add(L,R,l,mid,o<<1,c);
        if(R>mid)
            add(L,R,mid+1,r,o<<1|1,c);
        pushup(o);
    }
    int query(int L,int R,int l,int r,int o){
        if(L<=l&&r<=R){
            return seg[o];
        }
        int mid=(l+r)>>1,ans=0;
        pushdown(o,l,r);
        if(L<=mid)
            ans+=query(L,R,l,mid,o<<1);
        if(R>mid)
            ans+=query(L,R,mid+1,r,o<<1|1);
        return ans;
    }
};
namespace Seg3{//vi*i^2
    int seg[MAXN<<2]={},tag[MAXN<<2]={};
    void pushup(int o){
        seg[o]=seg[o<<1]+seg[o<<1|1];
    }
    int f(int x){
        return (2*x+1)*(x+1)*x/6;
    }
    int sum(int l,int r){
        return f(r)-f(l-1);
    }
    void build(int l,int r,int o,int *a){
        if(l==r){
            seg[o]=a[l]*l*l;
            return;
        }
        int mid=(l+r)>>1;
        build(l,mid,o<<1,a);
        build(mid+1,r,o<<1|1,a);
        pushup(o);
    }
    void pushdown(int o,int lx,int rx){
        if(tag[o]){
            int mid=(lx+rx)>>1;
            seg[o<<1]+=tag[o]*sum(lx,mid);
            seg[o<<1|1]+=tag[o]*sum(mid+1,rx);
            tag[o<<1]+=tag[o];
            tag[o<<1|1]+=tag[o];
            tag[o]=0;
        }
    }    
    void add(int L,int R,int l,int r,int o,int c){
        if(L<=l&&r<=R){
            seg[o]+=c*sum(l,r);
            tag[o]+=c;
            return;
        }
        int mid=(l+r)>>1;
        pushdown(o,l,r);
        if(L<=mid)
            add(L,R,l,mid,o<<1,c);
        if(R>mid)
            add(L,R,mid+1,r,o<<1|1,c);
        pushup(o);
    }
    int query(int L,int R,int l,int r,int o){
        if(L<=l&&r<=R){
            return seg[o];
        }
        int mid=(l+r)>>1,ans=0;
        pushdown(o,l,r);
        if(L<=mid)
            ans+=query(L,R,l,mid,o<<1);
        if(R>mid)
            ans+=query(L,R,mid+1,r,o<<1|1);
        return ans;
    }
    void debug(int l,int r,int o){
        printf("%lld %lld %lld %lld %lld\n",l,r,o,seg[o],tag[o]);
        if(l==r)
            return;
        int mid=(l+r)>>1;
        debug(l,mid,o<<1);
        debug(mid+1,r,o<<1|1);
    }
};
int gcd(int a,int b){
    return (b==0)?a:gcd(b,a%b);
}
int a[MAXN]={0},n,m;
using namespace Seg1;
using namespace Seg2;
using namespace Seg3;
signed main(){
    scanf("%lld %lld",&n,&m);
    n--;
    Seg1::build(1,n,1,a);
    Seg2::build(1,n,1,a);
    Seg3::build(1,n,1,a);
    for(int i=1;i<=m;i++){
        char c=getchar();
        while(c!='C'&&c!='Q')
            c=getchar();
        if(c=='C'){
            int l,r,val;
            scanf("%lld %lld %lld",&l,&r,&val);
            r--;
            Seg1::add(l,r,1,n,1,val);
            Seg2::add(l,r,1,n,1,val);
            Seg3::add(l,r,1,n,1,val);
            //Seg3::debug(1,n,1);
        }
        else{
            int l,r;
            scanf("%lld %lld",&l,&r);
            r--;
            int sum1=Seg1::query(l,r,1,n,1);
            int sum2=Seg2::query(l,r,1,n,1);
            int sum3=Seg3::query(l,r,1,n,1);
            //Seg3::debug(1,n,1);
            int ans=sum1*(r-l+1-r*l)+(r+l)*sum2-sum3;
            // printf("sum1=%lld sum2=%lld sum3=%lld %lld\n",sum1,sum2,sum3,ans);
            int t=(r-l+1)*(r-l+2)/2;
            int Gcd=gcd(t,ans);
            printf("%lld/%lld\n",ans/Gcd,t/Gcd);

        }
    }
    return 0;
}

标签：Lmid,sum,mid,高速公路,RL,HAOI2012,ans,pushdown,P2221
来源： https://www.cnblogs.com/dreagonm/p/10495200.html