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FZU 2082(过路费)

题目链接:传送门

题目大意:中文题,略

题目思路:树链剖分(注意要把边上的权值转移到深度较大的点上来维护)

          最后当top[x]==top[y]注意id[x]+1因为是维护的点而题目是边

          如果不+可能会出现重复加的情况。

  1 #include <iostream>  2 #include <cstdio>  3 #include <cstdlib>  4 #include <cmath>  5 #include <algorithm>  6 #include <cstring>  7 #include <stack>  8 #include <cctype>  9 #include <queue> 10 #include <string> 11 #include <vector> 12 #include <set> 13 #include <map> 14 #include <climits> 15 #define lson rt<<1,l,mid 16 #define rson rt<<1|1,mid+1,r 17 #define fi first 18 #define se second 19 #define ping(x,y) ((x-y)*(x-y)) 20 #define mst(x,y) memset(x,y,sizeof(x)) 21 #define mcp(x,y) memcpy(x,y,sizeof(y)) 22 using namespace std; 23 #define gamma 0.5772156649015328606065120 24 #define MOD 1000000007 25 #define inf 0x3f3f3f3f 26 #define N 50005 27 #define maxn 30010 28 typedef pair<int,int> PII; 29 typedef long long LL; 30 LL read(){ 31     LL x=0,f=1;char ch=getchar(); 32     while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();} 33     while(ch>=0&&ch<=9){x=(x<<3)+(x<<1)+ch-0;ch=getchar();} 34     return x*f; 35 } 36 int head[N],hcnt; 37 int n,m,flag,L,R; 38 int son[N],siz[N],fa[N],top[N]; 39 int id[N],tid,dep[N],posi[N]; 40 LL seg[N<<2]; 41 struct Node{int to,nxt;LL v;}node[N<<1]; 42 struct Edge{int x,y;LL v;}edge[N]; 43 void dfs1(int u,int f,int deep){ 44     fa[u]=f,dep[u]=deep,siz[u]=1; 45     for(int i=head[u];~i;i=node[i].nxt){ 46         int e=node[i].to;if(e==f)continue; 47         dfs1(e,u,deep+1);siz[u]+=siz[e]; 48         if(!son[u]||siz[son[u]]<siz[e])son[u]=e; 49     } 50 } 51 void dfs2(int u,int tp){ 52     top[u]=tp,id[u]=++tid,posi[tid]=u; 53     if(!son[u])return;dfs2(son[u],tp); 54     for(int i=head[u];~i;i=node[i].nxt){ 55         int e=node[i].to; 56         if(!id[e])dfs2(e,e); 57     } 58 } 59 LL query(int rt,int l,int r){ 60     if(L<=l&&r<=R)return seg[rt]; 61     int mid=l+r>>1;LL temp=0; 62     if(L<=mid)temp+=query(lson); 63     if(R>mid) temp+=query(rson); 64     return temp; 65 } 66 void update(int rt,int l,int r,LL v){ 67     if(l==r){seg[rt]=v;return;} 68     int mid=l+r>>1; 69     if(L<=mid)update(lson,v); 70     else update(rson,v); 71     seg[rt]=seg[rt<<1]+seg[rt<<1|1]; 72 } 73 void lca(int x,int y){ 74     LL res=0; 75     while(top[x]!=top[y]){ 76         if(dep[top[x]]<dep[top[y]])swap(x,y); 77         L=id[top[x]],R=id[x]; 78         res+=query(1,1,n); 79         x=fa[top[x]]; 80     } 81     if(dep[x]<dep[y])swap(x,y); 82     L=id[y]+1,R=id[x];   ///注意L +1 83     if(x!=y)res+=query(1,1,n); 84     printf("%lld\n",res); 85 } 86 void init(){ 87     mst(head,-1);hcnt=tid=0; 88     mst(siz,0);mst(son,0);mst(id,0); 89 } 90 int main(){ 91     int i,j,group,x,y,v,Case=0; 92     while(scanf("%d%d",&n,&m)!=EOF){ 93         init(); 94         for(i=1;i<n;++i){ 95             x=read(),y=read(),v=read(); 96             edge[i].x=x,edge[i].y=y,edge[i].v=v; 97             node[hcnt].v=v,node[hcnt].to=y,node[hcnt].nxt=head[x],head[x]=hcnt++; 98             node[hcnt].v=v,node[hcnt].to=x,node[hcnt].nxt=head[y],head[y]=hcnt++; 99         }100         dfs1(1,1,1);dfs2(1,1);101         for(i=1;i<n;++i){102             if(dep[edge[i].x]<dep[edge[i].y])swap(edge[i].x,edge[i].y);103             L=id[edge[i].x];104             update(1,1,n,edge[i].v);105         }106         while(m--){107             x=read();108             if(x==0){109                 x=read(),y=read();110                 L=id[edge[x].x];111                 update(1,1,n,y);112             }113             else{114                 x=read(),y=read();115                 lca(x,y);116             }117         }118     }119     return 0;120 }

 

FZU 2082(过路费)