hdu5044 Tree
Tree
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 845 Accepted Submission(s): 162
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
Source
思路：先求lca，对于点的修改，对于 u,v，假设 lca等于 u ,那么add[v] += val ;add[fa[u]] -= val ;
如果lca不为 u,v那么 ......看代码吧，
最后从根节点开始 up , add[fa] += add[son] ; 这个add[u]，就是u 增加的值
对于边的修改，我们先转为有根树，然后把边的权值分给两个端点中，深度高的那个，然后
就可以和点修改差不多了，具体看代码吧
这题真吭，很和谐的树链分割过不了。。。。
#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define INF 999999
#define mod 20140518
#define maxn 100010
using namespace std;
int head[maxn],next1[maxn*2],to[maxn*2] ;
int top ,id[maxn] ,n ,num[maxn],dep[maxn];
int xx[maxn],yy[maxn],cc[maxn],e[maxn];
int fa[maxn];
LL ans[maxn],add1[maxn],add2[maxn];
int out[maxn];
bool vi[maxn] ;
struct node
{
int v,id;
node(){}
node(int x,int y )
{
v = x ;id = y ;
}
};
vector<node>qe[maxn];
void Unit(int u,int v )
{
next1[top] = head[u] ;to[top] = v ;
head[u] = top++;
}
int find(int x )
{
if(x != fa[x])
fa[x] = find(fa[x]) ;
return fa[x];
}
void dfs(int u ,int f)
{
fa[u]=u;
vi[u] = true;
for(int i = head[u] ; i != -1 ; i = next1[i])
{
int v = to[i] ;
if(v==f) continue;
dfs(v,u) ;
fa[v] = u ;
}
node a;
for(int i = 0 ; i < qe[u].size();i++)
{
a = qe[u][i] ;
if(!vi[a.v]) continue ;
num[a.id] = find(a.v) ;
}
}
void bfs(int s)
{
memset(vi,0,sizeof(vi)) ;
memset(out,0,sizeof(out));
queue<int>q ;
q.push(s) ;
dep[s]=0;
vi[s] = true ;
while(!q.empty())
{
int u = q.front() ;
q.pop() ;
for(int i = head[u] ; i != -1 ; i = next1[i])
{
int v = to[i] ;
if(vi[v]) continue ;
out[u]++;
vi[v] = true ;
dep[v]=u;
id[v] = i/2+1 ;
q.push(v) ;
}
}
}
int next_int() {
char ch;
int res;
bool neg;
while (ch = getchar(), !isdigit(ch) && ch != '-')
;
if (ch == '-') {
neg = true;
res = 0;
} else {
neg = false;
res = ch - '0';
}
while (ch = getchar(), isdigit(ch))
res = res * 10 + ch - '0';
return neg ? -res : res;
}
int main()
{
char a[10] ;
int m,x,y,i,j,u,v;
int T,c ,case1=0;
cin >> T ;
while( T-- )
{
scanf("%d%d",&n,&m) ;
for(i = 0 ; i <= n ;i++)qe[i].clear();
memset(head,-1,sizeof(head)) ;
top=0;
for( i = 1 ; i < n ;i++)
{
x = next_int();
y = next_int();
Unit(x,y) ;
Unit(y,x) ;
}
bfs(1) ;
for( i = 1 ; i <= m ;i++)
{
scanf("%s",a) ;
x = next_int();
y = next_int();
c = next_int();
qe[x].push_back(node(y,i));
qe[y].push_back(node(x,i));
xx[i]=x;yy[i]=y;
cc[i]=c;
if(a[3]=='1')
{
e[i]=1;
}
else
{
e[i]=0;
}
}
memset(vi,0,sizeof(vi)) ;
memset(add1,0,sizeof(add1));
memset(add2,0,sizeof(add2));
dfs(1,-1);
int f;
for( i = 1 ; i <= m ;i++)
{
f = num[i];
u=xx[i];
v=yy[i];
if(e[i])
{
if(u==v)
{
add1[u] += cc[i];
add1[dep[f]] -= cc[i];
}
else{
add1[u] += cc[i];
add1[v] += cc[i];
add1[dep[f]] -= cc[i];
add1[f] -= cc[i];
}
}
else{
if(u==v)
{
add2[u] += cc[i];
add2[f] -= cc[i];
}
else{
add2[u] += cc[i];
add2[v] += cc[i];
add2[f] -= 2*cc[i];
}
}
}
queue<int>q;
for( i = 1 ; i <= n ;i++)if(!out[i])
{
q.push(i);
}
memset(vi,0,sizeof(vi));
while(!q.empty())
{
u = q.front();q.pop();
v = dep[u] ;
if(v==0) continue ;
add1[v] += add1[u] ;
add2[v] += add2[u] ;
out[v]--;
if(out[v]==0)q.push(v);
}
printf("Case #%d:\n",++case1);
bool flag=false;
for( i = 1 ; i <= n ;i++)
{
if(flag)printf(" ") ;
flag=true;
printf("%I64d",add1[i]);
}
puts("");
flag=false;
for( i = 2 ; i <= n ;i++)
{
ans[id[i]] = add2[i];
}
for( i = 1 ; i <n ;i++)
{
if(flag) printf(" ") ;
flag=true;
printf("%I64d",ans[i]);
}
puts("");
}
return 0 ;
}
