# 3307: 雨天的尾巴

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 843 Solved: 345
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### Description

N个点，形成一个树状结构。有M次发放，每次选择两个点x,y

20 50

8 6

10 6

18 6

20 10

7 20

2 18

19 8

1 6

14 20

16 10

13 19

3 14

17 18

11 19

4 11

15 14

5 18

9 10

12 15

11 14 87

12 1 87

14 3 84

17 2 36

6 5 93

17 6 87

10 14 93

5 16 78

6 15 93

15 5 16

11 8 50

17 19 50

5 4 87

15 20 78

1 17 50

20 13 87

7 15 22

16 11 94

19 8 87

18 3 93

13 13 87

2 1 87

2 6 22

5 20 84

10 12 93

18 12 87

16 10 93

8 17 93

14 7 36

7 4 22

5 9 87

13 10 16

20 11 50

9 16 84

10 17 16

19 6 87

12 2 36

20 9 94

9 2 84

14 1 94

5 5 94

8 17 16

12 8 36

20 17 78

12 18 50

16 8 94

2 19 36

10 18 36

14 19 50

4 12 50

87

36

84

22

87

87

22

50

84

87

50

36

87

93

36

94

16

87

50

50

1<=N,M<=100000

1<=a,b,x,y<=N

1<=z<=10^9

# 题解

http://blog.csdn.net/ww140142/article/details/48660513

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <cmath>
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define abs(x) ((x) < 0 ? -1 * (x) : (x))
template <class T>
inline void swap(T &x, T &y)
{
T tmp = x;x = y, y = tmp;
}
template <class T>
{
x = 0;char ch = getchar(), c = ch;
while(ch < '0' || ch > '9') c = ch, ch = getchar();
while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;
const int MAXN = 100000 + 10;

int n, q, ans[MAXN];

//离散化
int u[MAXN], v[MAXN], w2[MAXN], w1[MAXN], cnt[MAXN], tt, num[MAXN];
bool cmp(int a, int b)
{
return w1[a] < w1[b];
}

//LCA
struct Edge
{
int v,nxt;
Edge(int _v, int _nxt){v = _v;nxt = _nxt;}
Edge(){}
}edge[MAXN << 1];
int head[MAXN], cntt, tag[MAXN], p[MAXN][25], deep[MAXN], pos[MAXN], dfn[MAXN], t, M;
inline void insert(int a, int b)
{
}
void dfs(int x)
{
dfn[x] = ++ t;pos[t] = x;
for(int pos = head[x];pos;pos = edge[pos].nxt)
{
int v = edge[pos].v;
if(p[x][0] == v) continue;
p[v][0] = x, deep[v] = deep[x] + 1, dfs(v);
}
}
void yuchuli()
{
for(int i = 1;i <= n;++ i) if(!dfn[i]) dfs(i);
M = 0;
while((1 << M) <= n) ++ M; -- M;
for(int i = 1;i <= M;++ i)
for(int j = 1;j <= n;++ j)
p[j][i] = p[p[j][i - 1]][i - 1];
}
int LCA(int x, int y)
{
if(deep[x] < deep[y]) swap(x, y);
for(int i = M;i >= 0;-- i)
if(deep[x] - deep[y] >= (1 << i))
x = p[x][i];
if(x == y) return x;
for(int i = M;i >= 0;-- i)
if(p[x][i] != p[y][i])
x = p[x][i], y = p[y][i];
return p[x][0];
}

//SGT
struct Node
{
int data, p, ls, rs;
}node[MAXN * 50];
int stack[MAXN * 50], top, tot, root[MAXN];
inline int newnode()
{
}
inline void delnode(int x)
{
node[x].data = node[x].ls = node[x].rs = node[x].p = 0;
stack[++ top] = x;
}
inline void pushup(int o)
{
int ls = node[o].ls, rs = node[o].rs;
if(node[ls].data >= node[rs].data) node[o].data = node[ls].data, node[o].p = node[ls].p;
else node[o].data = node[rs].data, node[o].p = node[rs].p;
if(node[o].data == 0) node[o].p = 0;
return;
}
int merge(int x, int y, int l = 1, int r = tt)
{
if(!x) return y;
if(!y) return x;
int mid = (l + r) >> 1;
int tmp = newnode();
if(l == r)
{
node[tmp].data = node[x].data + node[y].data;
node[tmp].p = l;
}
else
{
node[tmp].ls = merge(node[x].ls, node[y].ls, l, mid);
node[tmp].rs = merge(node[x].rs, node[y].rs, mid + 1, r);
pushup(tmp);
}
delnode(x), delnode(y);
if(node[tmp].data == 0) node[tmp].p = 0;
return tmp;
}
void modify(int p, int k, int &o, int l = 1, int r = tt)
{
if(!o) o = newnode();
if(l == r && l == p)
{
node[o].data += k, node[o].p = l;
return;
}
int mid = (l + r) >> 1;
if(p <= mid) modify(p, k, node[o].ls, l, mid);
else modify(p, k, node[o].rs, mid + 1, r);
pushup(o);
}
int main()
{
for(int i = 1;i < n;++ i)
{
int tmp1, tmp2;
insert(tmp1, tmp2), insert(tmp2, tmp1);
}
yuchuli();
for(int i = 1;i <= q;++ i)
std::sort(cnt + 1, cnt + 1 + q, cmp);
num[++ tt] = w1[cnt[1]], w2[cnt[1]] = tt;
for(int i = 1;i < q;++ i)
if(w1[cnt[i]] != w1[cnt[i + 1]]) num[++ tt] = w1[cnt[i + 1]] , w2[cnt[i + 1]] = tt;
else w2[cnt[i + 1]] = tt;
for(int i = 1;i <= q;++ i)
{
int lca = LCA(u[i], v[i]);
modify(w2[i], 1, root[u[i]]);
modify(w2[i], 1, root[v[i]]);
modify(w2[i], -1, root[lca]);
if(p[lca][0]) modify(w2[i], -1, root[p[lca][0]]);
}
for(int j = n;j >= 1;-- j)
{
int i = pos[j];
for(int pos = head[i];pos;pos = edge[pos].nxt)
{
int v = edge[pos].v;
if(v == p[i][0]) continue;
root[i] = merge(root[i], root[v]);
}
ans[i] = node[root[i]].p;
}
for(int i = 1;i <= n;++ i) printf("%d\n", num[ans[i]]);
return 0;
}

posted @ 2018-02-24 09:35  嘒彼小星  阅读(243)  评论(0编辑  收藏  举报