思路

考慮每一條邊的貢獻，然后推式子
\[ \begin{align}&\sum_{i}V_i\times(R-i+1)\times(i-L+1)\\=&\sum_{i}V_i\left[(Ri-i^2+i)-(RL-iL+L)+(R-i+1)\right]\\=&\sum_{i}V_i\left[Ri-i^2+i-RL+Li-L+R-i+1\right]\\=&\sum_{i}Vi\left[(Ri+Li)-i^2-RL+(R-L+1)\right]\\=&\sum_{i}(Ri+Li)V_i-\sum_{i}i^2V_i-\sum_{i}RLV_i+\sum_{i}(R-L+1)V_i\\=&(R+L)\sum_{i}iVi-\sum_{i}i^2V_i-RL\sum_{i}V_i+(R-L+1)\sum_{i}V_i\\=&(R+L)\sum_iiV_i-\sum_ii^2V_i+(R-L+1-RL)\sum_{i}V_i\end{align} \]
然后用線段樹維護就好了

代碼

#include <cstdio> #include <algorithm> #include <cstring> #define int long long using namespace std; const int MAXN = 100100; namespace Seg1{//v_iint 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*iint 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^2int 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; }

轉載于:https://www.cnblogs.com/dreagonm/p/10495200.html

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