Description
Input
Output
Sample Input
4 3 5
1 9 2
7 6 5 1
2 3
3 1
3 4
1 2 3 2
1 1 2
1 4 2
0 2 1
1 1 2
1 4 2
1 9 2
7 6 5 1
2 3
3 1
3 4
1 2 3 2
1 1 2
1 4 2
0 2 1
1 1 2
1 4 2
Sample Output
84
131
27
84
131
27
84
HINT
Solution
还是改成括号序的写法吧……感觉好理解还好码……
一开始插入和删除函数是像下面这样分开写的,$vis$函数加加减减不知道哪里错了……如果有大爷看出来哪里错了和我说一声啊QAQ
1 void Ins(int x) 2 { 3 vis[x]++; 4 if (vis[x]==2) Delicacy-=v[c[x]]*w[Keg[c[x]]], Keg[c[x]]--; 5 else Keg[c[x]]++, Delicacy+=v[c[x]]*w[Keg[c[x]]]; 6 } 7 8 void Del(int x) 9 { 10 vis[x]--; 11 if (vis[x]==1) Keg[c[x]]++, Delicacy+=v[c[x]]*w[Keg[c[x]]]; 12 else Delicacy-=v[c[x]]*w[Keg[c[x]]], Keg[c[x]]--; 13 } 14 15 void Recov(int x,int val) 16 { 17 if (vis[x]==1) Del(x), c[x]=val, Ins(x); 18 c[x]=val; 19 }
后来改成天下第一的写法对$vis$搞异或就过了……至今不知道为什么……
还有我才发现我二轮的树上莫队写的是对的啊……只不过$cmp$函数写错真的太开心了……
虽然多那30分也无济于事就是了……
话说我二轮的树上莫队既然是对的那我2月份写的那一发树分块糖果公园为什么T成狗啊
Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<algorithm> 5 #define N (200009) 6 #define LL long long 7 using namespace std; 8 9 struct Que{int l,r,t,num,flag; LL ans;}Q[N]; 10 struct Mdf{int pre,nxt,pos;}M[N]; 11 struct Edge{int to,next;}edge[N<<1]; 12 LL Delicacy,v[N],w[N]; 13 int n,m,q,x,y,opt,l=1,r,flag,Q_num,M_num,Time,dfs_num; 14 int Keg[N*5],vis[N],ID[N],a[N]; 15 int Fir[N],Sec[N],f[N][18],Depth[N],c[N]; 16 int head[N],num_edge; 17 bool cmp1(Que a,Que b) 18 { 19 if (ID[a.l]==ID[b.l]) 20 return ID[a.r]==ID[b.r]?a.t<b.t:ID[a.r]<ID[b.r]; 21 else return ID[a.l]<ID[b.l]; 22 } 23 bool cmp2(Que a,Que b) {return a.num<b.num;} 24 25 void add(int u,int v) 26 { 27 edge[++num_edge].to=v; 28 edge[num_edge].next=head[u]; 29 head[u]=num_edge; 30 } 31 32 void Ins(int x) 33 { 34 if (vis[x]) Delicacy-=v[c[x]]*w[Keg[c[x]]], Keg[c[x]]--; 35 else Keg[c[x]]++, Delicacy+=v[c[x]]*w[Keg[c[x]]]; 36 vis[x]^=1; 37 } 38 39 void Recov(int x,int val) 40 { 41 if (vis[x]) Ins(x),c[x]=val,Ins(x); 42 else c[x]=val; 43 } 44 45 void MoQueue(int num) 46 { 47 while (Time<Q[num].t) Recov(M[Time+1].pos,M[Time+1].nxt), Time++; 48 while (Time>Q[num].t) Recov(M[Time].pos,M[Time].pre), Time--; 49 while (l<Q[num].l) Ins(a[l++]); 50 while (l>Q[num].l) Ins(a[--l]); 51 while (r<Q[num].r) Ins(a[++r]); 52 while (r>Q[num].r) Ins(a[r--]); 53 Q[num].ans=Delicacy; 54 if (Q[num].flag) 55 { 56 Ins(Q[num].flag); 57 Q[num].ans=Delicacy; 58 Ins(Q[num].flag); 59 } 60 } 61 62 void DFS(int x,int fa) 63 { 64 f[x][0]=fa; 65 for (int i=1; i<=17; ++i) f[x][i]=f[f[x][i-1]][i-1]; 66 Depth[x]=Depth[fa]+1; Fir[x]=++dfs_num; 67 a[dfs_num]=x; 68 for (int i=head[x]; i; i=edge[i].next) 69 if (edge[i].to!=fa) DFS(edge[i].to,x); 70 Sec[x]=++dfs_num; a[dfs_num]=x; 71 } 72 73 int LCA(int x,int y) 74 { 75 if (Depth[x]<Depth[y]) swap(x,y); 76 for (int i=17; i>=0; --i) 77 if (Depth[f[x][i]]>=Depth[y]) x=f[x][i]; 78 if (x==y) return x; 79 for (int i=17; i>=0; --i) 80 if (f[x][i]!=f[y][i]) x=f[x][i], y=f[y][i]; 81 return f[x][0]; 82 } 83 int main() 84 { 85 scanf("%d%d%d",&n,&m,&q); 86 int unit=pow(n,2.0/3.0); 87 for (int i=1; i<=(n*2); ++i) ID[i]=(i-1)/unit+1; 88 for (int i=1; i<=m; ++i) scanf("%lld",&v[i]); 89 for (int i=1; i<=n; ++i) scanf("%lld",&w[i]); 90 for (int i=1; i<=n-1; ++i) 91 scanf("%d%d",&x,&y), add(x,y), add(y,x); 92 for (int i=1; i<=n; ++i) scanf("%d",&c[i]); 93 DFS(1,0); 94 for (int i=1; i<=q; ++i) 95 { 96 scanf("%d%d%d",&opt,&x,&y); 97 if (opt==1) 98 { 99 if (Fir[x]>Fir[y]) swap(x,y); 100 int lca=LCA(x,y); 101 if (lca==x) x=Fir[x], y=Fir[y], flag=0; 102 else x=Sec[x], y=Fir[y], flag=lca; 103 Q[++Q_num]=(Que){x,y,M_num,i,flag,0}; 104 } 105 else M[++M_num]=(Mdf){c[x],y,x}, c[x]=y; 106 } 107 for (int i=M_num; i>=1; --i) 108 c[M[i].pos]=M[i].pre; 109 sort(Q+1,Q+Q_num+1,cmp1); 110 for (int i=1; i<=Q_num; ++i) 111 MoQueue(i); 112 sort(Q+1,Q+Q_num+1,cmp2); 113 for (int i=1; i<=Q_num; ++i) 114 printf("%lld ",Q[i].ans); 115 }