题意：
一棵树 支持删边加边、路径权值加值、路径权值改值、路径求第二大的数字和其个数
思路：
LCT的第二题 题意已经把功能都告诉了 比较裸
要注意的是权值加值和改值两个操作的标记下放问题 要先down改值 再down加值
对于路径的操作通过mroot变换树的形态再access拿出路径比较方便 不要像我上一篇一样搞lca
代码：
#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define N 100010 #define L(x) (ch[x][0]) #define R(x) (ch[x][1]) #define inf -2147483640 int ch[N][2], pre[N], rev[N], me[N]; int cov[N], add[N], size[N], max1[N], num1[N], max2[N], num2[N]; bool rt[N]; struct helper { int val, num; bool operator<(const helper ff) const { return val > ff.val; } } hel[10]; void Update_COV(int u, int d) { if (!u) return; me[u] = d; cov[u] = d; add[u] = inf; max1[u] = d; num1[u] = size[u]; max2[u] = inf; num2[u] = 0; } void Update_Add(int u, int d) { if (!u) return; me[u] += d; if (add[u] != inf) add[u] += d; else add[u] = d; max1[u] += d; if (max2[u] != inf) max2[u] += d; } void Update_Rev(int u) { if (!u) return; swap(L(u), R(u)); rev[u] ^= 1; } void down(int u) { if (rev[u]) { Update_Rev(L(u)); Update_Rev(R(u)); rev[u] = 0; } if (cov[u] != inf) { Update_COV(L(u), cov[u]); Update_COV(R(u), cov[u]); cov[u] = inf; } if (add[u] != inf) { Update_Add(L(u), add[u]); Update_Add(R(u), add[u]); add[u] = inf; } } void up(int u) { size[u] = size[L(u)] + size[R(u)] + 1; hel[0].val = max1[L(u)]; hel[0].num = num1[L(u)]; hel[1].val = max2[L(u)]; hel[1].num = num2[L(u)]; hel[2].val = max1[R(u)]; hel[2].num = num1[R(u)]; hel[3].val = max2[R(u)]; hel[3].num = num2[R(u)]; hel[4].val = me[u]; hel[4].num = 1; sort(hel, hel + 5); int i; for (i = 1; i < 5; i++) { if (hel[i].val != hel[i - 1].val) break; } max1[u] = hel[0].val; max2[u] = hel[i].val; num1[u] = num2[u] = 0; for (i = 0; i < 5; i++) { if (hel[i].val == max1[u]) num1[u] += hel[i].num; else if (hel[i].val == max2[u]) num2[u] += hel[i].num; } } //Rotate P Splay 一般不变 void Rotate(int x) { int y = pre[x], kind = ch[y][1] == x; ch[y][kind] = ch[x][!kind]; pre[ch[y][kind]] = y; pre[x] = pre[y]; pre[y] = x; ch[x][!kind] = y; if (rt[y]) rt[y] = false, rt[x] = true; else ch[pre[x]][ch[pre[x]][1] == y] = x; up(y); } //P函数先将splay根结点到u的路径上所有的结点的标记逐级下放 void P(int u) { if (!rt[u]) P(pre[u]); down(u); } void Splay(int u) { P(u); while (!rt[u]) { int fa = pre[u], ffa = pre[fa]; if (rt[fa]) Rotate(u); else if ((R(ffa) == fa) == (R(fa) == u)) Rotate(fa), Rotate(u); else Rotate(u), Rotate(u); } up(u); } //将root到u的路径变成实边 int Access(int u) { int v = 0; for (; u; u = pre[v = u]) { Splay(u); rt[R(u)] = true, rt[R(u) = v] = false; up(u); } return v; } //使u成为它所在的树的根 void mroot(int u) { Access(u); Splay(u); Update_Rev(u); } //连接两棵树 u接在v上 void link(int u, int v) { mroot(u); pre[u] = v; } //u-v边断开 void cut(int u, int v) { mroot(u); Access(u); Splay(v); pre[L(v)] = pre[v]; pre[v] = 0; rt[L(v)] = true; L(v) = 0; up(v); } //u-v路径权值变为w void COV(int u, int v, int w) { mroot(u); Access(v); Splay(v); Update_COV(v, w); } //u-v路径+w void ADD(int u, int v, int w) { mroot(u); Access(v); Splay(v); Update_Add(v, w); } //u-v路径最大值 void query(int u, int v) { mroot(u); Access(v); Splay(v); if (max2[v] != inf) printf("%d %d\n", max2[v], num2[v]); else printf("ALL SAME\n"); } struct edge { int v, next; } ed[N * 2]; int head[N], tot; void addedge(int u, int v) { ed[tot].v = v; ed[tot].next = head[u]; head[u] = tot++; } //利用dfs初始化pre数组 建立LCT void dfs(int u) { for (int i = head[u]; ~i; i = ed[i].next) { int v = ed[i].v; if (pre[v] != 0) continue; pre[v] = u; dfs(v); } } int main() { int T, t, n, m, i, op, u, v, x, y, w; max1[0] = max2[0] = inf; rt[0] = true; scanf("%d", &T); for (t = 1; t <= T; t++) { scanf("%d%d", &n, &m); for (i = 1; i <= n; i++) { scanf("%d", &me[i]); max1[i] = me[i]; max2[i] = inf; num1[i] = 1; num2[i] = 0; size[i] = 1; add[i] = cov[i] = inf; L(i) = R(i) = pre[i] = rev[i] = 0; rt[i] = true; } tot = 0; memset(head, -1, sizeof(head)); for (i = 1; i < n; i++) { scanf("%d%d", &u, &v); addedge(u, v); addedge(v, u); } pre[1] = -1; dfs(1); pre[1] = 0; printf("Case #%d:\n", t); while (m--) { scanf("%d", &op); if (op == 1) { scanf("%d%d%d%d", &u, &v, &x, &y); cut(u, v); link(x, y); } else if (op == 2) { scanf("%d%d%d", &u, &v, &w); COV(u, v, w); } else if (op == 3) { scanf("%d%d%d", &u, &v, &w); ADD(u, v, w); } else { scanf("%d%d", &u, &v); query(u, v); } } } return 0; }
