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  • 【LG3320】[SDOI2015]寻宝游戏

    【LG3320】[SDOI2015]寻宝游戏

    题面

    洛谷

    题解

    不需要建虚树的虚树2333。。。

    贪心地想一下,起始节点肯定是在关键点上,访问顺序就是(dfs)序。

    那么对于每次询问,

    [Ans=dis(S_1,S_s)+sum_{i=1}^{s-1}dis(S_i,S_{i+1}) ]

    (set)维护一下就好了

    代码

    #include <iostream> 
    #include <cstdio> 
    #include <cstdlib> 
    #include <cstring> 
    #include <cmath> 
    #include <algorithm>
    #include <set> 
    using namespace std; 
    inline int gi() { 
        register int data = 0, w = 1; 
        register char ch = 0; 
        while (!isdigit(ch) && ch != '-') ch = getchar(); 
        if (ch == '-') w = -1, ch = getchar(); 
        while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
        return w * data; 
    }
    typedef long long ll; 
    const int MAX_N = 1e5 + 5; 
    struct Graph { int to, next, cost; } e[MAX_N << 1]; 
    int fir[MAX_N], e_cnt; 
    void clearGraph() { memset(fir, -1, sizeof(fir)); } 
    void Add_Edge(int u, int v, int w) { e[e_cnt] = (Graph){v, fir[u], w}; fir[u] = e_cnt++; }
    int N, M;
    ll dis[MAX_N], ans; 
    int dep[MAX_N], size[MAX_N], top[MAX_N], son[MAX_N], fa[MAX_N], dfn[MAX_N], tim; 
    void dfs1(int x) { 
    	size[x] = 1, dep[x] = dep[fa[x]] + 1; 
    	for (int i = fir[x]; ~i; i = e[i].next) {
    		int v = e[i].to; if (v == fa[x]) continue;
    		dis[v] = dis[x] + e[i].cost; 
    		fa[v] = x, dfs1(v), size[x] += size[v]; 
    		if (size[v] > size[son[x]]) son[x] = v; 
    	} 
    } 
    void dfs2(int x, int tp) { 
    	top[x] = tp, dfn[x] = ++tim; 
    	if (son[x]) dfs2(son[x], tp); 
    	for (int i = fir[x]; ~i; i = e[i].next) {
    		int v = e[i].to; if (v == fa[x] || v == son[x]) continue; 
    	    dfs2(v, v); 
    	} 
    } 
    int LCA(int u, int v) { 
    	while (top[u] != top[v]) {
    		if (dep[top[u]] < dep[top[v]]) swap(u, v); 
    		u = fa[top[u]]; 
    	} 
    	return dep[u] < dep[v] ? u : v; 
    } 
    ll DIS(int x, int y) { return dis[x] + dis[y] - 2ll * dis[LCA(x, y)]; }
    struct cmp { bool operator () (int x, int y) const { return dfn[x] < dfn[y]; } } ; 
    set<int, cmp> s = set<int, cmp>(cmp());
    set<int, cmp> :: iterator ite; 
    bool vis[MAX_N]; 
    int main () { 
    #ifndef ONLINE_JUDGE 
        freopen("cpp.in", "r", stdin); 
    #endif 
    	clearGraph(); 
    	N = gi(), M = gi(); 
    	for (int i = 1; i < N; i++) { 
    		int u = gi(), v = gi(), w = gi(); 
    		Add_Edge(u, v, w), Add_Edge(v, u, w); 
    	} 
    	dfs1(1), dfs2(1, 1); 
    	while (M--) { 
    		int x = gi(); 
    		if (vis[x]) { 
    			ite = s.find(x); 
    			if (s.size() < 3) { ans = 0; goto P1; } 
    			int pre, nxt; 
    			if (ite == s.begin()) pre = *--s.end(); 
    			else pre = *(--ite), ++ite; 
    			if (ite == --s.end()) nxt = *s.begin(); 
    			else nxt = *(++ite), --ite; 
    			ans -= DIS(pre, x) + DIS(x, nxt) - DIS(pre, nxt); 
    		  P1 :
    			s.erase(ite); 
    		} else { 
    			ite = s.insert(x).first; 
    			if (s.size() == 1) goto P2; 
    			int pre, nxt; 
    			if (ite == s.begin()) pre = *--s.end(); 
    			else pre = *(--ite), ++ite; 
    			if (ite == --s.end()) nxt = *s.begin(); 
    			else nxt = *(++ite), --ite; 
    			ans += DIS(pre, x) + DIS(x, nxt) - DIS(pre, nxt); 
    		  P2: ; 
    		} 
    		vis[x] ^= 1;
    		printf("%lld
    ", ans); 
    	} 
    	return 0; 
    } 
    
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  • 原文地址:https://www.cnblogs.com/heyujun/p/10360190.html
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