异象石

发表于 更新于 分类于

Problem’s Website

  • 异象石

    Solution

  • 首先我们可以得出这是一颗树，我们可以用一些关于树的基本操作，比如我们可以求出节点的dfs序，根据dfs序升序排序，把异象石的节点首尾相连，累加相邻节点的路径长度，得到的结果正好是答案的两倍，可以画图观察一下。
  • 那么我们可以用set来维护dfs序，用LCA来计算路径长度，就可以把这道题A掉。

    Code

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    24
    25
    26
    27
    28
    29
    30
    31
    32
    33
    34
    35
    36
    37
    38
    39
    40
    41
    42
    43
    44
    45
    46
    47
    48
    49
    50
    51
    52
    53
    54
    55
    56
    57
    58
    59
    60
    61
    62
    63
    64
    65
    66
    67
    68
    69
    70
    71
    72
    73
    74
    75
    76
    77
    78
    79
    80
    81
    82
    83
    84
    85
    86
    87
    88
    89
    90
    91
    92
    93
    94
    95
    96
    97
    98
    99
    100
    101
    102
    103
    104
    105
    106
    107
    108
    109
    110
    111
    112
    113
    114
    115
    116
    117
    118
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<set>
    #define gc getchar()
    #define ll long long
    #define Int set <int> :: iterator
    using namespace std;
    int sc() {
    	int xx = 0, ff = 1; char cch = gc;
    	while (cch < '0' || cch > '9') {
    		if (cch == '-') ff = -1; cch = gc;
    	}
    	while (cch >= '0' && cch <= '9') {
    		xx = (xx << 1) + (xx << 3) + (cch ^ '0'); cch = gc;
    	}
    	return xx * ff;
    }
    struct TREE {
    	int next, to, dis;
    }tree[100010 << 1];
    int n, cnt, m, id;
    int f[100010][21], df[100010], rdf[100010], deep[100010];
    ll d[100010][21]; 
    ll ans;
    set <int> s;
    int head[100010 << 1];
    void ADD(int from, int to, int dis) {
    	tree[++cnt].next = head[from];
    	tree[cnt].to = to;
    	tree[cnt].dis = dis;
    	head[from] = cnt;
    }
    void D_F(int son, int fa) {
    	for(int i = 0; i <= 19; i++) {
    		f[son][i + 1] = f[f[son][i]][i];
    		d[son][i + 1] = d[son][i] + d[f[son][i]][i];
    	}
    	for(int i = head[son]; i; i = tree[i].next) {
    		int to = tree[i].to, dis = tree[i].dis;
    		if(to == fa) continue;
    		f[to][0] = son;
    		d[to][0] = dis;
    		deep[to] = deep[son] + 1;
    		D_F(to, son);
    	}
    }
    ll LCA(int x, int y) {
    	ll res = 0;
    	if(deep[x] > deep[y]) swap(x, y);
    	ll de = deep[y] - deep[x];
    	for(int i = 0; i <= 20; i++)
    		if(de & (1 << i))
    			res += d[y][i], y = f[y][i];
    	if(x == y) return res;
    	for(int i = 20; i >= 0; i--) {
    		if(f[x][i] != f[y][i]) {
    			res += d[x][i] + d[y][i];
    			x = f[x][i]; y = f[y][i];
    		}
    	}
    	return res + d[x][0] + d[y][0];
    }
    void dfs(int now) {
    	df[now] = ++id;
    	rdf[id] = now;
    	for(int i = head[now]; i; i = tree[i].next) {
    		int to = tree[i].to;
    		if(!df[to])
    			dfs(to);
    	}
    }
    Int L(Int now) {
    	if(now == s.begin()) return --s.end();
    	return --now;
    }
    Int R(Int now) {
    	if(now == --s.end()) return s.begin();
    	return ++now;
    }
    int main() {
    	n = sc();
    	for(int i = 1; i < n; i++) {
    		int x = sc(), y = sc(), z = sc();
    		ADD(x, y, z); ADD(y, x, z);
    	}
    	D_F(1, 0);
    	dfs(1);
    	m = sc();
    	while(m--) {
    		char c[2];
    		cin >> c[0];
    		if(c[0] == '?') {
    			printf("%lld\n", (ans >> 1));
    		}
    		else if(c[0] == '+') {
    			int ID = sc();
    			Int now;
    			if(s.size()) {
    				now = s.lower_bound(df[ID]);
    				if(now == s.end()) now = s.begin();
    				int first = *L(now);
    				ans += LCA(ID, rdf[first]) + LCA(ID, rdf[*now]) - LCA(rdf[first], rdf[*now]);
    			}
    			s.insert(df[ID]);
    		}
    		else if(c[0] == '-') {
    			int ID = sc();
    			Int now = s.find(df[ID]);
    			int first = *L(now);
    			now = R(now);
    			ans -= LCA(ID, rdf[first]) + LCA(ID, rdf[*now]) - LCA(rdf[first], rdf[*now]);
    			s.erase(df[ID]);
    		}	
    	}
    	return 0;
    }

rp++