Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4217 Accepted Submission(s): 1164
3 2 5 1 2 3 2 1 2 3 I 1 3 5 Q 2 D 1 2 2 Q 1 Q 3
7 4 8Hint1.The number of enemies may be negative. 2.Huge input, be careful.
题意:给一棵树,并给定各个点权的值,然后有3种操作:
I C1 C2 K: 把C1与C2的路径上的所有点权值加上K
D C1 C2 K:把C1与C2的路径上的所有点权值减去K
Q C:查询节点编号为C的权值
大胆地用C++提交吧!
#include<map> #include<string> #include<cstring> #include<cstdio> #include<cstdlib> #include<cmath> #include<queue> #include<vector> #include<iostream> #include<algorithm> #include<bitset> #include<climits> #include<list> #include<iomanip> #include<stack> #include<set> using namespace std; int head[50010],tail; struct Edge { int to,next; }edge[50000*2+10]; void add(int from,int to) { edge[tail].to=to; edge[tail].next=head[from]; head[from]=tail++; } int siz[50010],dep[50010],fa[50010],son[50010]; void dfs1(int from) { siz[from]=1; for(int i=head[from];i!=-1;i=edge[i].next) { int to=edge[i].to; if(to!=fa[from]) { fa[to]=from; dep[to]=dep[from]+1; dfs1(to); siz[from]+=siz[to]; if(siz[to]>siz[son[from]]) son[from]=to; } } } int w[50010],top[50010],cnt; void dfs2(int from,int tp) { w[from]=++cnt; top[from]=tp; if(son[from]==0) return; dfs2(son[from],tp); for(int i=head[from];i!=-1;i=edge[i].next) { int to=edge[i].to; if(to!=son[from]&&to!=fa[from]) dfs2(to,to); } } struct Tree { int l,r,m,v; }tree[50000*4+10]; void build(int l,int r,int k) { tree[k].l=l; tree[k].r=r; tree[k].m=l+r>>1; tree[k].v=0; if(l==r) return; build(l,tree[k].m,k<<1); build(tree[k].m+1,r,k<<1|1); } void update(int l,int r,int val,int k) { if(l==tree[k].l&&r==tree[k].r) { tree[k].v+=val; return; } if(tree[k].v!=0) { tree[k<<1].v+=tree[k].v; tree[k<<1|1].v+=tree[k].v; tree[k].v=0; } if(r<=tree[k].m) update(l,r,val,k<<1); else if(l>tree[k].m) update(l,r,val,k<<1|1); else { update(l,tree[k].m,val,k<<1); update(tree[k].m+1,r,val,k<<1|1); } } int seek(int x,int k) { if(tree[k].l==tree[k].r&&tree[k].l==x) return tree[k].v; if(tree[k].v!=0) { tree[k<<1].v+=tree[k].v; tree[k<<1|1].v+=tree[k].v; tree[k].v=0; } if(x<=tree[k].m) return seek(x,k<<1); return seek(x,k<<1|1); } void update(int x,int y,int val) { while(top[x]!=top[y]) { if(dep[top[x]]<dep[top[y]]) swap(x,y); update(w[top[x]],w[x],val,1); x=fa[top[x]]; } int l=w[x],r=w[y]; if(l>r) swap(l,r); update(l,r,val,1); } int num[50010]; char od[2]; int main() { int n,m,p; while(scanf("%d%d%d",&n,&m,&p)!=EOF) { for(int i=1;i<=n;i++) scanf("%d",num+i); tail=0; memset(head,-1,sizeof(head)); while(m--) { int from,to; scanf("%d%d",&from,&to); add(from,to); add(to,from); } memset(son,0,sizeof(son)); dfs1(1); cnt=0; dfs2(1,1); build(1,cnt,1); for(int i=1;i<=n;i++) update(i,i,num[i]); while(p--) { scanf("%s",od); if(od[0]=='Q') { int x; scanf("%d",&x); printf("%d\n",seek(w[x],1)); } else { int l,r,v; scanf("%d%d%d",&l,&r,&v); if(od[0]=='D') v=-v; update(l,r,v); } } } }
原文:http://blog.csdn.net/stl112514/article/details/44758921