HDU 5957 Query on a graph (拓扑 + bfs序 + 树剖 + 线段树)

题意：一个图有n个点，n条边，定义D(u,v)为u到v的距离，S(u,k)为所有D(u,v)<=k的节点v的集合 有m次询问(0<=k<=2):

1 u k d：将集合S(u,k)的所有节点的权值加d

2 u k:询问集合S(u,k)的所有节点的权值之和

析：把这个图树成两部分，一个是一个环，然后剩下的森林。

这个环可以用拓扑来求，看这个博客吧，讲的非常细了。

http://blog.csdn.net/qq_31759205/article/details/75049074

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int maxm = 1e6 + 5;
const int mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Edge{
  int to, next;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
int in[maxn];

void addEdge(int u, int v){
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
  ++in[v];
}

void topsort(){
  queue<int> q;
  for(int i = 1; i <= n; ++i)  if(in[i] == 1)  q.push(i);
  while(!q.empty()){
    int u = q.front();  q.pop();
    for(int i = head[u]; ~i; i = edge[i].next){
      int v = edge[i].to;
      if(in[v] > 1){
        --in[v];
        if(in[v] == 1)  q.push(v);
      }
    }
  }
}

int ch[maxn][2], p[maxn], fp[maxn], pos;
int fa[maxn];
int sonL[maxn], sonR[maxn], ssonL[maxn], ssonR[maxn];

void bfs(int s){
  queue<int> q;
  q.push(s);

  while(!q.empty()){
    int u = q.front();  q.pop();
    sonL[u] = ssonL[u] = INF;
    sonR[u] = ssonR[u] = 0;
    for(int i = head[u]; ~i; i = edge[i].next){
      int v = edge[i].to;
      if(in[v] > 1 || fa[u] == v)  continue;
      p[v] = ++pos;
      fp[pos] = v;
      fa[v] = u;
      sonL[u] = min(sonL[u], p[v]);
      sonR[u] = max(sonR[u], p[v]);
      q.push(v);
    }
    ssonL[fa[u]] = min(ssonL[fa[u]], sonL[u]);
    ssonR[fa[u]] = max(ssonR[fa[u]], sonR[u]);
  }
}

LL sum[maxn<<2], addv[maxn<<2];
inline void push_up(int rt){ sum[rt] = sum[rt<<1] + sum[rt<<1|1]; }
inline void push_down(int rt, int len){
  int l = rt<<1, r = rt<<1|1;
  sum[l] += (len -  (len>>1)) * addv[rt];
  sum[r] += (len>>1) * addv[rt];
  addv[l] += addv[rt];
  addv[r] += addv[rt];
  addv[rt] = 0;
}

void update(int L, int R, int val, int l, int r, int rt){
  if(L > R)   return ;
  if(L <= l && r <= R){
    sum[rt] += (LL)(r - l + 1) * val;
    addv[rt] += val;
    return ;
  }
  if(addv[rt])  pd(rt, r - l + 1);
  int m = l + r >> 1;
  if(L <= m)  update(L, R, val, lson);
  if(R > m)   update(L, R, val, rson);
  pu(rt);
}

LL query(int L, int R, int l, int r, int rt){
  if(L > R)  return 0LL;
  if(L <= l && r <= R)  return sum[rt];
  if(addv[rt])  pd(rt, r - l + 1);
  int m = l + r >> 1;
  LL ans = 0;
  if(L <= m)  ans = query(L, R, lson);
  if(R > m)  ans += query(L, R, rson);
  return ans;
}

void praper(){
  topsort();
  for(int u = 1; u <= n; ++u)  if(in[u] > 1){
    int j = 0;
    for(int i = head[u]; ~i; i = edge[i].next){
      int v = edge[i].to;
      if(in[v] > 1)  ch[u][j++] = v;
    }
    p[u] = ++pos;
    fp[pos] = u;
    fa[u] = 0;
    bfs(u);
  }
}

void init(){
  ms(head, -1);  cnt = 0;  ms(in, 0);
  pos = 0;  ms(sum, 0);  ms(addv, 0);
}

void update(int u, int k, int val){
  int f = fa[u];
  if(k == 0)  update(p[u], p[u], val, all);  // self
  else if(k == 1){
    if(in[u] > 1){  // on the circle
      update(p[ch[u][0]], p[ch[u][0]], val, all);  // left neighbor
      update(p[ch[u][1]], p[ch[u][1]], val, all);  // right neighbor
    }
    else  // not on the circle
      update(p[f], p[f], val, all);  // its father
    update(sonL[u], sonR[u], val, all); // left son - right son
    update(p[u], p[u], val, all);  //self
  }
  else {
    if(in[u] > 1){  // on the circle
      int vv[2], cnt = 0;
      for(int i = 0; i < 2; ++i){
        int v = ch[u][i];
        update(p[v], p[v], val, all);  // its neighbor
        update(sonL[v], sonR[v], val, all);  // its neighbor's sons
        for(int j = 0; j < 2; ++j){
          int t = ch[v][j];
          if(t == u || t == ch[u][i^1])  continue;  // insure not calculate twice times
          if(cnt == 1 && vv[0] == t)  continue;
          vv[cnt++] = t;
        }
      }
      for(int i = 0; i < cnt; ++i)
        update(p[vv[i]], p[vv[i]], val, all);  // its neighbor's neighbor
      update(p[u], p[u], val, all);  // self take care  the below have no
    }
    else{  // not on the circle
      update(p[f], p[f], val, all);  // its father
      update(sonL[f], sonR[f], val, all);  // its father's son
      if(in[f] > 1){  // its father is on the circle
        update(p[ch[f][0]], p[ch[f][0]], val, all);  // its father's left neighbor
        update(p[ch[f][1]], p[ch[f][1]], val, all);  // its father's fight neighbor
      }
      else  // its father is not on the circle
        update(p[fa[f]], p[fa[f]], val, all);  // its father's father
    }
    update(sonL[u], sonR[u], val, all);  // sons
    update(ssonL[u], ssonR[u], val, all);  // ssons
  }
}

LL query(int u, int k){  // the same as update
  int f = fa[u];
  LL ans = 0;
  if(k == 0)  return query(p[u], p[u], all);
  else if(k == 1){
    ans = query(sonL[u], sonR[u], all) + query(p[u], p[u], all);
    if(in[u] == 1)  ans += query(p[f], p[f], all);
    else ans += query(p[ch[u][0]], p[ch[u][0]], all) + query(p[ch[u][1]], p[ch[u][1]], all);
  }
  else{
    ans = query(sonL[u], sonR[u], all) + query(ssonL[u], ssonR[u], all);
    if(in[u] > 1){
      ans += query(p[u], p[u], all);
      int vv[2], cnt = 0;
      for(int i = 0; i < 2; ++i){
        int v = ch[u][i];
        ans += query(p[v], p[v], all);
        ans += query(sonL[v], sonR[v], all);
        for(int j = 0; j < 2; ++j){
          int t = ch[v][j];
          if(t == u || t == ch[u][i^1])  continue;
          if(cnt == 1 && t == vv[0])  continue;
          vv[cnt++] = t;
        }
      }
      for(int i = 0; i < cnt; ++i)
        ans += query(p[vv[i]], p[vv[i]], all);
    }
    else {
      ans += query(p[f], p[f], all) + query(sonL[f], sonR[f], all);
      if(in[f] > 1)  ans += query(p[ch[f][0]], p[ch[f][0]], all) + query(p[ch[f][1]], p[ch[f][1]], all);
      else ans += query(p[fa[f]], p[fa[f]], all);
    }
  }
  return ans;
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    init();
    for(int i = 0; i < n; ++i){
      int u, v;
      scanf("%d %d", &u, &v);
      addEdge(u, v);
      addEdge(v, u);
    }
    praper();
    scanf("%d", &m);
    char op[10];
    int u, k, d;
    while(m--){
      scanf("%s %d %d", op, &u, &k);
      if(op[0] == 'M'){
        scanf("%d", &d);
        update(u, k, d);
      }
      else printf("%I64d\n", query(u, k));
    }
  }
  return 0;
}