• [洛谷P4949]最短距离


    题目大意:给一棵基环树,两种操作:

    1. $1;x;y:$把第$x$条边长度改成$y$
    2. $2;x;y:$查询$x$到$y$的最短距离

    题解:发现最短距离只有两种可能,第一个是树上的距离,第二种是经过多出来的一条边,都求出来比较一下就行了。修改时,若修改的是多出来的边,直接修改即可,若不是可以把它子树中所有点的距离修改,可以按$dfs$序变成区间修改

    卡点:

    C++ Code:

    #include <cstdio>
    #include <algorithm>
    #define maxn 100010
    int head[maxn], cnt = 1;
    struct Edge {
    	int to, nxt, w;
    } e[maxn << 1];
    inline void add(int a, int b, int c) {
    	e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
    	e[++cnt] = (Edge) {a, head[b], c}; head[b] = cnt;
    }
    
    int n, m, circle;
    int l[maxn], r[maxn], w[maxn];
    
    #define M 17
    int fa[maxn][M], dep[maxn], sz[maxn];
    int dfn[maxn], idx;
    bool vis[maxn];
    void dfs(int u, int father = 0) {
    	for (int i = 1; i < M; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
    	vis[u] = true;
    	dfn[u] = ++idx;
    	sz[u] = 1;
    	for (int i = head[u]; i; i = e[i].nxt) if (i ^ father ^ 1) {
    		int v = e[i].to;
    		if (!vis[v]) {
    			*fa[v] = u;
    			dep[v] = dep[u] + 1;
    			dfs(v, i);
    			sz[u] += sz[v];
    		} else circle = i >> 1;
    	}
    }
    inline int LCA(int x, int y) {
    	if (x == y) return x;
    	if (dep[x] < dep[y]) std::swap(x, y);
    	for (int i = dep[x] - dep[y]; i; i &= i - 1) x = fa[x][__builtin_ctz(i)];
    	if (x == y) return x;
    	for (int i = M - 1; ~i; i--) if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
    	return *fa[x];
    }
    #undef M
    
    int last[maxn];
    namespace BIT {
    	long long Tr[maxn], res;
    	inline void add(int p, int num) {for (; p <= n; p += p & -p) Tr[p] += num;}
    	inline int ask(int p) {for (res = 0; p; p &= p - 1) res += Tr[p]; return res;}
    }
    
    inline void __modify(int u, int v, int w) {BIT::add(u, w), BIT::add(v + 1, -w);}
    inline void modify(int u, int v, int w) {
    	if (dfn[u] > dfn[v]) std::swap(u, v);
    	__modify(dfn[v], dfn[v] + sz[v] - 1, w - ::last[v]); ::last[v] = w;
    }
    
    inline long long dis(int x, int y) {return BIT::ask(dfn[x]) + BIT::ask(dfn[y]) - BIT::ask(dfn[LCA(x, y)]) * 2;}
    inline long long solve(int x, int y) {
    	long long ans = dis(x, y);
    	ans = std::min(ans, dis(x, l[circle]) + dis(y, r[circle]) + w[circle]);
    	return std::min(ans, dis(x, r[circle]) + dis(y, l[circle]) + w[circle]);
    }
    
    int main() {
    	scanf("%d%d", &n, &m);
    	for (int i = 1; i <= n; i++) {
    		scanf("%d%d%d", l + i, r + i, w + i);
    		add(l[i], r[i], w[i]);
    	}
    	dfs(1);
    	for (int i = 1; i <= n; i++) if (i != circle) modify(l[i], r[i], w[i]);
    	while (m --> 0) {
    		int op, x, y;
    		scanf("%d%d%d", &op, &x, &y);
    		if (op == 1) {
    			if (x == circle) w[circle] = y;
    			else modify(l[x], r[x], y);
    		} else printf("%lld
    ", solve(x, y));
    	}
    	return 0;
    }
    

      

  • 相关阅读:
    ES6入门 阮一峰
    NPM
    移动端BUG
    配置每次git push 不需要输入账号密码
    移动端rem布局,用户调整手机字体大小或浏览器字体大小后导致页面布局出错问题
    课程表
    岛屿数量
    二叉树的右视图
    c++设计模式——工厂模式
    克隆图
  • 原文地址:https://www.cnblogs.com/Memory-of-winter/p/9870090.html
Copyright © 2020-2023  润新知