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  • 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
    ", query(u, k));
        }
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7688480.html
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