Aragorn's StoryTime Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17156 Accepted Submission(s): 4524 Problem Description Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time. Input Multiple test cases, process to the end of input. Output For each query, you need to output the actually number of enemies in the specified camp. Sample Input 3 2 5 1 2 3 2 1 2 3 I 1 3 5 Q 2 D 1 2 2 Q 1 Q 3 Sample Output 7 4 8Hint 1.The number of enemies may be negative. 2.Huge input, be careful.Source 2011 Multi-University Training Contest 13 - Host by HIT Recommend We have carefully selected several similar problems for you: 3964 3965 3962 3963 3967 |
【题意】
有一棵树n点m=n-1边,每个点有个权值。有三种操作:“I”/“D” u v k 对u到v的路径上所有点的权值 增加/减少 k。“Q” u 查询u点当前值。
【分析】
树链剖分模板即可。(这题完全可以再增加点趣味性,查询u到v路径上的点权和)
我分别用线段树和树状数组维护。线段树耗时大概是树状数组的两倍。
【代码-线段树】
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX=5e4+5;
const int INF=0x3f3f3f3f;
/*********** 图 ***********/
struct node{
int t,next;
}edge[MAX*2];
int head[MAX],cnt;
void init(int n)
{
cnt=0; memset(head,-1,sizeof(head[0])*(n+1));
}
void addedge(int u,int v)
{
edge[cnt]=node{v,head[u]};
head[u]=cnt++;
}
/************ 树链剖分 **************/
int dep[MAX],dad[MAX],siz[MAX],son[MAX];
int top[MAX],rk[MAX],id[MAX];
void dfs1(int u,int pre)
{
siz[u]=1; //u本身
son[u]=-1;
int maxsiz=0; //最大的儿子子树大小
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].t;
if(v==pre)continue;
dep[v]=dep[u]+1;
dad[v]=u;
dfs1(v,u); //注意dfs的位置
siz[u]+=siz[v];
if(maxsiz<siz[v])
{
maxsiz=siz[v];
son[u]=v;
}
}
}
void dfs2(int u,int fat,int &tag) //结点u,重链顶端
{
top[u]=fat;
rk[u]=++tag; //时间戳
id[tag]=u; //tag对应在树上的结点号
if(son[u]==-1)return; //没有孩子
dfs2(son[u],fat,tag); //优先重链dfs
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].t;
if(v==dad[u]||v==son[u])continue;
dfs2(v,v,tag);
}
}
void cuttree(int root) //将root为根的树链剖分
{
int tag=0;
dep[root]=0; //根的深度和父结点做一个初始化
dad[root]=root; //这是一个假设,根的父结点还是根
dfs1(root,root);
dfs2(root,root,tag);
}
/************* 线段树 ****************/
ll sum[MAX<<2],mark[MAX<<2];
void build(int root,int l,int r) //初始线段树
{
if(l==r)//叶子节点
{
mark[root]=0;
sum[root]=0;
return;
}
int mid=(l+r)/2;
build(root*2,l,mid); //左子树的创建
build(root*2+1,mid+1,r); //右子树
sum[root]=sum[root*2]+sum[root*2+1];
mark[root]=0;
}
void nodeupdate(int root,int l,int r,ll num)
{
mark[root] +=num; //=
sum[root]+=num*(r-l+1); //=
}
void pushdown(int root,int l,int r)//传递给两孩子
{
if(mark[root]==0)return;
int mid=(l+r)/2;
nodeupdate(root*2,l,mid,mark[root]);
nodeupdate(root*2+1,mid+1,r,mark[root]);
mark[root]=0;
}
void update(int left,int right,ll num, int root,int l,int r)//区间[kl,kr]修改
{
if(left<=l&&r<=right){
nodeupdate(root,l,r,num);
return;
}
pushdown(root,l,r);
int mid=(l+r)/2;
if(left<=mid)
update(left,right,num,root*2,l,mid);
if(mid<right)
update(left,right,num,root*2+1,mid+1,r);
sum[root]=sum[root*2]+sum[root*2+1];
}
ll Qsum(int left,int right, int root,int l,int r)//区间和
{
if(left<=l&&r<=right)
return sum[root];
pushdown(root,l,r);
int mid=(l+r)/2;
ll res=0;
if(left<=mid)
res+=Qsum(left,right,root*2,l,mid);//区间在左子树
if(right>mid)
res+=Qsum(left,right,root*2+1,mid+1,r);//在右子树
return res;
}
/************** 树链->线段树 *********/
void modify(int u,int v,ll num, int limit) //修改
{
int fu=top[u],fv=top[v];
while(fu!=fv) //uv不在同一条链上
{
if(dep[fu]>=dep[fv]) //u的深度大
{
update(rk[fu],rk[u],num,1,1,limit);
u=dad[fu]; fu=top[u];
}
else
{
update(rk[fv],rk[v],num,1,1,limit);
v=dad[fv]; fv=top[v];
}
} //循环结束时,fu-fv在同一链
if(dep[u]>dep[v])swap(u,v);
update(rk[u],rk[v],num,1,1,limit);
}
ll query(int u,int v, int limit) //查询u-v路径
{
ll res=0;
int fu=top[u],fv=top[v];
while(fu!=fv) //uv不在同一条链上
{
if(dep[fu]>=dep[fv]) //u的深度大
{
res+=Qsum(rk[fu],rk[u],1,1,limit);
u=dad[fu]; fu=top[u];
}
else
{
res+=Qsum(rk[fv],rk[v],1,1,limit);
v=dad[fv]; fv=top[v];
}
} //循环结束时,fu-fv在同一链
if(dep[u]>dep[v])swap(u,v);
res+=Qsum(rk[u],rk[v],1,1,limit);
return res;
}
/************** solve **************/
int a[MAX];
bool solve()
{
int n,m,P,u,v,k;
if(scanf("%d%d%d",&n,&m,&P)==EOF)return 0;
init(n); //初始图
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
for(int i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
cuttree(1); //树链剖分
build(1,1,n); //建立线段树
for(int i=1;i<=n;i++)modify(i,i,a[i],n);
char op[3];
while(P--)
{
scanf("%s",op);
if(op[0]=='Q')
{
scanf("%d",&u);
ll ans=query(u,u,n); //查询
printf("%lld\n",ans);
}
else
{
scanf("%d%d%d",&u,&v,&k);
if(op[0]=='D')k=-k;
modify(u,v,k,n);
}
}
return 1;
}
int main()
{
while(solve());
}
【代码-树状数组】
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX=5e4+5;
const int INF=0x3f3f3f3f;
/*********** 图 ***********/
struct node{
int t,next;
}edge[MAX*2];
int head[MAX],cnt;
void init(int n)
{
cnt=0; memset(head,-1,sizeof(head[0])*(n+1));
}
void addedge(int u,int v)
{
edge[cnt]=node{v,head[u]};
head[u]=cnt++;
}
/************ 树链剖分 **************/
int dep[MAX],dad[MAX],siz[MAX],son[MAX];
int top[MAX],rk[MAX],id[MAX];
void dfs1(int u,int pre)
{
siz[u]=1; //u本身
son[u]=-1;
int maxsiz=0; //最大的儿子子树大小
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].t;
if(v==pre)continue;
dep[v]=dep[u]+1;
dad[v]=u;
dfs1(v,u); //注意dfs的位置
siz[u]+=siz[v];
if(maxsiz<siz[v])
{
maxsiz=siz[v];
son[u]=v;
}
}
}
void dfs2(int u,int fat,int &tag) //结点u,重链顶端
{
top[u]=fat;
rk[u]=++tag; //时间戳
id[tag]=u; //tag对应在树上的结点号
if(son[u]==-1)return; //没有孩子
dfs2(son[u],fat,tag); //优先重链dfs
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].t;
if(v==dad[u]||v==son[u])continue;
dfs2(v,v,tag);
}
}
void cuttree(int root) //将root为根的树链剖分
{
int tag=0;
dep[root]=0; //根的深度和父结点做一个初始化
dad[root]=root; //这是一个假设,根的父结点还是根
dfs1(root,root);
dfs2(root,root,tag);
}
/************* bit tree ****************/
ll bit[MAX]; //区间增减+单点查询
void add(int k,ll num,int limit)
{
for(;k<=limit;k+=k&-k)
bit[k]+=num;
}
ll read(int k)
{
ll res=0;
for(;k;k-=k&-k)
res+=bit[k];
return res;
}
void update(int left,int right,ll num,int limit)
{
add(left,num,limit);
add(right+1,-num,limit);
}
/************** 树链-> bit tree *********/
void modify(int u,int v,ll num,int limit) //修改uv之间都增加num
{
int fu=top[u],fv=top[v];
while(fu!=fv) //uv不在同一条链上
{
if(dep[fu]>=dep[fv]) //u的深度大
{
update(rk[fu],rk[u],num,limit);
u=dad[fu]; fu=top[u];
}
else
{
update(rk[fv],rk[v],num,limit);
v=dad[fv]; fv=top[v];
}
} //循环结束时,fu fv在同一链
if(dep[u]>dep[v])swap(u,v);
update(rk[u],rk[v],num,limit);
}
/************** solve **************/
int a[MAX];
bool solve()
{
int n,m,P,u,v,k;
if(scanf("%d%d%d",&n,&m,&P)==EOF)return 0;
init(n); //初始图
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
for(int i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
cuttree(1); //树链剖分
for(int i=1;i<=n;i++)bit[i]=0;
for(int i=1;i<=n;i++)update(rk[i],rk[i],a[i],n);
char op[3];
while(P--)
{
scanf("%s",op);
if(op[0]=='Q')
{
scanf("%d",&u);
ll ans=read(rk[u]); //单点查询
printf("%lld\n",ans);
}
else
{
scanf("%d%d%d",&u,&v,&k);
if(op[0]=='D')k=-k;
modify(u,v,k,n);
}
}
return 1;
}
int main()
{
while(solve());
}