Total Submission(s): 1024 Accepted Submission(s): 188
Problem Description
You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N
There are N - 1 edges numbered from 1 to N - 1.
Each node has a value and each edge has a value. The initial value is 0.
There are two kind of operation as follows:
● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.
● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.
After finished M operation on the tree, please output the value of each node and edge.
There are N - 1 edges numbered from 1 to N - 1.
Each node has a value and each edge has a value. The initial value is 0.
There are two kind of operation as follows:
● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.
● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.
After finished M operation on the tree, please output the value of each node and edge.
Input
The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.
The first line of each case contains two integers N ,M (1 ≤ N, M ≤105),denoting the number of nodes and operations, respectively.
The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.
For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -105 ≤ k ≤ 105)
The first line of each case contains two integers N ,M (1 ≤ N, M ≤105),denoting the number of nodes and operations, respectively.
The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.
For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -105 ≤ k ≤ 105)
Output
For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.
The second line contains N integer which means the value of each node.
The third line contains N - 1 integer which means the value of each edge according to the input order.
The second line contains N integer which means the value of each node.
The third line contains N - 1 integer which means the value of each edge according to the input order.
Sample Input
2
4 2
1 2
2 3
2 4
ADD1 1 4 1
ADD2 3 4 2
4 2
1 2
2 3
1 4
ADD1 1 4 5
ADD2 3 2 4
Sample Output
Case #1:
1 1 0 1
0 2 2
Case #2:
5 0 0 5
0 4 0
//去年學長做的時候還能過，記得是G++跑過的，現在做就要c++然後手動開棧才跑了3000+ms，不知道是不是新加了數據。
#include<stdio.h>
#include<string.h>
#include<math.h>
#include <string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<cmath>
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 1<<29
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
#define EPS 1e-8
#define MID int mid=(l+r)>>1; int ls=o<<1; int rs=o<<1|1;
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int N=100005;
int father[N];
bool vis[N];
int op[N];
int a[N],b[N],lca[N];
long long v[N];
long long dis[N];
long long dis2[N];
int faa[N];
int out[N];
struct Edge{
int v,next,id;
}e[N*4];
int head[N],ask[N];
int cnt;
void add1(int u,int v,int id)
{
e[cnt].v=v;
e[cnt].next=head[u];
e[cnt].id=id;
head[u]=cnt++;
}
void add2(int u,int v,int id)
{
e[cnt].v=v;
e[cnt].next=ask[u];
e[cnt].id=id;
ask[u]=cnt++;
}
int Find(int x)
{
if(father[x]==x)return x;
return father[x]=Find(father[x]);
}
void LCA(int now,int fa)
{
faa[now]=fa;
father[now]=now;
vis[now]=true;
for(int i=ask[now];i!=-1;i=e[i].next){
int v=e[i].v;
int id=e[i].id;
if(vis[v])
lca[id]=Find(v);
}
for(int i=head[now];i!=-1;i=e[i].next){
int v=e[i].v;
int id=e[i].id;
if(v==fa) continue;
out[id]=v;
LCA(v,now);
father[v]=now;
}
}
void dfs(int now,int fa)
{
long long ret=0;
long long ret2=0;
for(int i=head[now];i!=-1;i=e[i].next){
int v=e[i].v;
if(v==fa)continue;
dfs(v,now);
ret+=dis[v];
ret2+=dis2[v];
}
dis[now]+=ret;
dis2[now]+=ret2;
}
int main()
{
int T;
int n,m;
int x,y;
char str[5];
int Case=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
memset(dis,0,sizeof(dis));
memset(dis2,0,sizeof(dis2));
memset(head,-1,sizeof(head));
memset(ask,-1,sizeof(ask));
cnt=0;
for(int i=0;i<n-1;i++){
scanf("%d%d",&x,&y);
add1(x,y,i);
add1(y,x,i);
}
for(int i=0;i<m;i++){
scanf("%s%d%d%I64d",str,&x,&y,&v[i]);
if(str[3]=='1')op[i]=1;
else op[i]=2;
a[i]=x;b[i]=y;
add2(x,y,i);
add2(y,x,i);
}
LCA(1,-1);
faa[1]=0;
for(int i=0;i<m;i++){
if(op[i]==2){
dis[a[i]]+=v[i];
dis[b[i]]+=v[i];
dis[lca[i]]-=v[i]*2;
}else {
dis2[a[i]]+=v[i];
dis2[b[i]]+=v[i];
dis2[lca[i]]-=v[i];
dis2[faa[lca[i]]]-=v[i];
}
}
dfs(1,-1);
printf("Case #%d:\n",Case++);
printf("%I64d",dis2[1]);
for(int i=2;i<=n;i++){
printf(" %I64d",dis2[i]);
}printf("\n");
if(n==1)
{printf("\n");continue;}
printf("%I64d",dis[out[0]]);
for(int i=1;i<n-1;i++){
printf(" %I64d",dis[out[i]]);
}printf("\n");
}
return 0;
}其中對點值的更新，一次操作
其中爲什麼最近公共祖先要減v，而它的父節點也要減v，從差分數列的角度考慮
要想從a[i]到a[j]都加上d，那麼b[i]+d,b[j+1]-d,那麼從x到y要想更新值，那麼就要將x到lca看成一個差分數列，把y到lca的父節點看陳給一個差分數列
如圖，那麼x到lca之前的那個點要更新，就執行
- dis2[b[i]]+=v[i];
- dis2[lca[i]]-=v[i];如果y到lca想更新，那麼就要y處加v，lca的爸爸減v，那麼從x到y的路上的點全都更新了
- 邊一個道理，但是邊不會更新兩次，就直接lca處減2*v就好了