HDU5044---Tree 树链剖分

大致题意：add1 u v   u到v路径上所有点的权值加上k，add2  u 到v路径上所有边的权值加上k

最后输出所有点的权值，边的权值。。树链剖分预处理然后来个线性O(n)的操作。刚开始用线段树tle了.

  1 #pragma comment(linker, "/STACK:1024000000,1024000000")

  2 #include <cstdio>

  3 #include <cstdlib>

  4 #include <cstring>

  5 #include <iostream>

  6 #include <algorithm>

  7 using namespace std;

  8 typedef unsigned long long ull;

  9 typedef long long ll;

 10 const int inf = 0x3f3f3f3f;

 11 const int maxn = 1e5+10;

 12 struct

 13 {

 14     int  to,next;

 15 } e[maxn<<1];

 16 int head[maxn],edge;

 17 void add(int x,int y)

 18 {

 19     e[edge].to = y;

 20     e[edge].next = head[x];

 21     head[x] = edge++;

 22 }

 23 

 24 int son[maxn],fa[maxn],siz[maxn],dep[maxn];

 25 void dfs1(int root)

 26 {

 27     siz[root] = 1;

 28     son[root] = 0;

 29     for (int i = head[root]; i > 0; i = e[i].next)

 30     {

 31         if (fa[root] != e[i].to)

 32         {

 33             dep[e[i].to] = dep[root] + 1;

 34             fa[e[i].to] = root;

 35             dfs1(e[i].to);

 36             if (siz[son[root]] < siz[e[i].to])

 37                 son[root] = e[i].to;

 38             siz[root] += siz[e[i].to];

 39         }

 40     }

 41 }

 42 int top[maxn],pos[maxn],fp[maxn],tot;

 43 void dfs2(int root,int f)

 44 {

 45     top[root] = f;

 46     pos[root] = tot++;

 47     fp[pos[root]] = root;

 48     if (son[root]>0)

 49         dfs2(son[root],top[root]);

 50     for (int i = head[root]; i > 0; i = e[i].next)

 51         if (fa[root] != e[i].to && e[i].to != son[root])

 52             dfs2(e[i].to,e[i].to);

 53 }

 54 ll addv[3][maxn<<2];

 55 int k;

 56 void pre_update(int ua,int ub,int cho)

 57 {

 58     int f1 = top[ua];

 59     int f2 = top[ub];

 60     while (f1 != f2)

 61     {

 62         if (dep[f1] < dep[f2])

 63             swap(f1,f2),swap(ua,ub);

 64         addv[cho][pos[f1]] += k;

 65         addv[cho][pos[ua]+1] -= k;

 66         ua = fa[f1];

 67         f1 = top[ua];

 68     }

 69     if (dep[ua] > dep[ub])

 70         swap(ua,ub);

 71     if (cho == 1)

 72         addv[cho][pos[ua]] += k,addv[cho][pos[ub]+1] -= k;

 73     if (cho == 2)

 74         addv[cho][pos[son[ua]]] += k,addv[cho][pos[ub]+1] -= k;

 75 }

 76 int n,m,d[maxn][2],link[maxn];

 77 void init()

 78 {

 79     scanf ("%d%d",&n,&m);

 80     int root = 1;

 81     dep[root] = fa[root] = 0;

 82     edge = tot = 1;

 83     memset(head,0,sizeof(head));

 84     memset(siz,0,sizeof(siz));

 85     memset(addv,0,sizeof(addv));

 86     for (int i = 1; i < n; i++)

 87     {

 88         int u,v;

 89         scanf ("%d%d",&u,&v);

 90         d[i][0] = u;

 91         d[i][1] = v;

 92         add(u,v),add(v,u);

 93     }

 94     dfs1(root);

 95     dfs2(root,root);

 96     for (int i = 1; i < n; i++)

 97     {

 98         if (dep[d[i][0]] < dep[d[i][1]])

 99             swap(d[i][0],d[i][1]);

100         link[d[i][0]] = i;

101     }

102 }

103 ll ans1[maxn],ans2[maxn];

104 int main(void)

105 {

106     //freopen("in.txt","r",stdin);

107     int t;

108     int cas = 1;

109     scanf ("%d",&t);

110     while (t--)

111     {

112         init();

113         for (int i = 0; i < m; i++)

114         {

115             char op[10];

116             int u,v;

117             scanf ("%s%d%d%d",op,&u,&v,&k);

118             pre_update(u,v,op[3]-'0');

119         }

120         for (int i = 1; i <= n; i++)

121         {

122             addv[1][i] += addv[1][i-1];

123             addv[2][i] += addv[2][i-1];

124             ans1[fp[i]] = addv[1][i];

125             ans2[link[fp[i]]] = addv[2][i];

126         }

127         printf("Case #%d:\n",cas++);

128         printf("%I64d",ans1[1]);

129         for (int i = 2; i <= n; i++)

130         {

131             printf(" %I64d",ans1[i]);

132         }

133         printf("\n");

134         if (n>1)

135             printf("%I64d",ans2[1]);

136         for (int i = 2; i < n; i++)

137         {

138             printf(" %I64d",ans2[i]);

139         }

140         printf("\n");

141     }

142     return 0;

143 }

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