hdu 5044树链剖分

Tree

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 475    Accepted Submission(s): 94

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.

 

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)

 

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.

 

Sample Input

24 21 22 32 4ADD1 1 4 1ADD2 3 4 24 21 22 31 4ADD1 1 4 5ADD2 3 2 4

 

Sample Output

Case #1:1 1 0 10 2 2Case #2:5 0 0 50 4 0

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

 

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。。还是太水了，比赛的时候点的标号和树链的标号弄混掉了，wa了无数遍。。还需要再熟练啊。

/***********************************************\ |Author: YMC |Created Time: 2014/9/22 21:23:55 |File Name: hdu 5029 self.cpp |Description: \***********************************************/#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>#include <algorithm>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define mset(l,n) memset(l,n,sizeof(l))#define rep(i,n) for(int i=0;i<n;++i)#define maxx(a) memset(a, 0x3f, sizeof(a))#define zero(a) memset(a, 0, sizeof(a))#define srep(i,n) for(int i = 1;i <= n;i ++)#define MP make_pairconst int inf=0x3f3f3f3f ;const double eps=1e-8 ;const double pi=acos (-1.0);typedef long long ll;using namespace std;#define maxn 100010vector <int> ma[maxn];int pathid[maxn],pathtop[maxn],dep[maxn];int que[maxn],s,t,fa[maxn],size[maxn],son[maxn],tot;int fq[maxn];ll an[maxn],bn[maxn];int n,m,u,v,cnt;int tp;inline void add_edge(int a,int b){    ma[a].push_back(b);ma[b].push_back(a);}//树链剖分void buildpath() {    int u,v;    s = 0;t = 0;    que[t ++] = 1;          //注意修改起点    fa[1] = -1;dep[1] = 1;    while(s < t) {        u = que[s ++];        rep(i,ma[u].size()) {            v = ma[u][i];            if(v != fa[u]) {                fa[v] = u; dep[v] = dep[u] + 1; que[t ++] = v;            }        }    }    for(int j = n-1;j >= 0;--j) {        u = que[j];        son[u] = -1; size[u] = 1;        rep(i,ma[u].size()) {            v = ma[u][i];            if(v != fa[u]) {                size[u] += size[v];                if(son[u] == -1 || size[v] > size[son[u]]) son[u] = v;            }        }        if(son[u] == -1) son[u] = u;    }    memset(pathtop,-1,sizeof(pathtop));    cnt = 1;    for(int i = 0;i<n;++i) {        u = que[i];        if(pathtop[u] != -1) continue;        int top = u;        for(;;) {            pathtop[u] = top;            pathid[u] = cnt ++;            fq[pathid[u]] = u;            if(son[u] == u) break;            u = son[u];        }    }}void init(int n){     rep(i,n+2){        ma[i].clear();    }    memset(an,0,sizeof(an));    memset(bn,0,sizeof(bn));}void change1(int u,int v,int z) {    int f1 = pathtop[u],f2 = pathtop[v];    while(f1 != f2) {        if(dep[f1] < dep[f2]) { //始终让f1在下面            swap(f1,f2);            swap(u,v);        }        an[pathid[f1]] += z;        an[pathid[u] + 1] -= z;        u = fa[f1];        f1 = pathtop[u];    }    if(dep[u] > dep[v]) swap(u,v);    an[pathid[u]] += z;    an[pathid[v] + 1] -= z;}void change2(int u,int v,ll z) {    //if(u == v) return;    int f1 = pathtop[u],f2 = pathtop[v];    while(f1 != f2) {        if(dep[f1] < dep[f2]) { //始终让f1在下面            swap(f1,f2);            swap(u,v);        }        bn[pathid[f1]] += z;        bn[pathid[u] + 1] -= z;        u = fa[f1];        f1 = pathtop[u];    }    if(u == v) return ;    if(dep[u] > dep[v]) swap(u,v);    bn[pathid[son[u]]] += z;    bn[pathid[v]+1] -= z;}char ch[20];ll ans[maxn];ll ans1[maxn];//int lll[maxn],rrr[maxn];pair<int,int> pa[maxn];int main() {    //freopen("input.txt","r",stdin);     int T;    scanf("%d",&T);    int cas = 1;    while(T--) {        scanf("%d %d",&n,&m);        init(n);        rep(i,n-1) {            scanf("%d %d",&u,&v);            add_edge(u,v);            pa[i] = make_pair(u,v);        }        buildpath();        //剖分        rep(i,n-1){            if(dep[pa[i].first] < dep[pa[i].second]) {                swap(pa[i].first,pa[i].second);            }        }        rep(i,m){            scanf("%s %d %d %d",ch,&u,&v,&tp);            if(ch[3] == '1'){                change1(u,v,tp);            }            else {                change2(u,v,tp);            }        }        printf("Case #%d:\n",cas ++);        ll sum = 0;        for(int i=1;i<=n;++i){            sum += an[i];            ans1[fq[i]] = sum;        }        for(int i=1;i<n;++i){            printf("%I64d ",ans1[i]);        }        printf("%I64d\n",ans1[n]);        if(n == 1){            puts("");            continue;        }        sum = 0;        for(int i=2;i<=n;++i){            sum += bn[i];            ans[fq[i]] = sum;        }        rep(i,n-2){            printf("%I64d ",ans[pa[i].first]);        }        printf("%I64d\n",ans[pa[n-2].first]);    }    return 0;}