首页 > 代码库 > POJ 3237 Tree 树链剖分
POJ 3237 Tree 树链剖分
单点替换,区间取相反数,区间求最大值,其实线段树里面只要保存最小值和最大值就可以了,取反之后最小值变成最大值最大值变成最小值。
#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <set>#include <bitset>#include <queue>#include <stack>#include <string>#include <iostream>#include <cmath>#include <climits>using namespace std;const int maxn = 10000 + 5;int n, head[maxn], nxt[maxn << 1], u[maxn << 1], v[maxn << 1], w[maxn << 1];int tid[maxn], maxson[maxn], fa[maxn], dep[maxn], siz[maxn], idcnt, ecnt;int top[maxn], maxv[maxn << 2], minv[maxn << 2], sval[maxn], lazy[maxn << 2];int eid[maxn << 1], rnum[maxn];#define lson rt << 1, l, mid#define rson rt << 1 | 1, mid + 1, rvoid init() { memset(head, -1, sizeof(head)); memset(maxson, -1, sizeof(maxson)); ecnt = idcnt = 0;}void adde(int uu, int vv, int ww, int id) { u[ecnt] = uu; v[ecnt] = vv; w[ecnt] = ww; eid[ecnt] = id; nxt[ecnt] = head[uu]; head[uu] = ecnt++;}void dfs1(int now, int nowfa, int nowdep) { fa[now] = nowfa; dep[now] = nowdep; siz[now] = 1; for(int i = head[now]; ~i; i = nxt[i]) if(v[i] != nowfa) { dfs1(v[i], now, nowdep + 1); siz[now] += siz[v[i]]; if(maxson[now] == -1 || siz[v[i]] > siz[maxson[now]]) { maxson[now] = v[i]; } }}void dfs2(int now, int tp) { top[now] = tp; tid[now] = ++idcnt; if(maxson[now] == -1) return; dfs2(maxson[now], tp); for(int i = head[now]; ~i; i = nxt[i]) { if(v[i] != fa[now] && v[i] != maxson[now]) { dfs2(v[i], v[i]); } }}void pushup(int rt) { minv[rt] = min(minv[rt << 1], minv[rt << 1 | 1]); maxv[rt] = max(maxv[rt << 1], maxv[rt << 1 | 1]);}void pushdown(int rt) { if(lazy[rt] == 0) return; int lc = rt << 1, rc = rt << 1 | 1; lazy[lc] ^= lazy[rt]; lazy[rc] ^= lazy[rt]; swap(minv[lc], maxv[lc]); swap(minv[rc], maxv[rc]); minv[lc] = -minv[lc]; minv[rc] = -minv[rc]; maxv[lc] = -maxv[lc]; maxv[rc] = -maxv[rc]; lazy[rt] = 0;}void build(int rt, int l, int r) { int mid = (l + r) >> 1; lazy[rt] = 0; if(l == r) { minv[rt] = maxv[rt] = sval[l]; } else { build(lson); build(rson); pushup(rt); }}void change(int rt, int l, int r, int pos, int val) { if(l == r) { minv[rt] = maxv[rt] = val; } else { int mid = (l + r) >> 1; pushdown(rt); if(pos <= mid) change(lson, pos, val); else change(rson, pos, val); pushup(rt); }}void update(int rt, int l, int r, int ql, int qr) { if(ql <= l && qr >= r) { swap(minv[rt], maxv[rt]); minv[rt] = -minv[rt]; maxv[rt] = -maxv[rt]; lazy[rt] ^= 1; } else { int mid = (l + r) >> 1; pushdown(rt); if(ql <= mid) update(lson, ql, qr); if(qr > mid) update(rson, ql, qr); pushup(rt); }}int query(int rt, int l, int r, int ql, int qr) { if(ql <= l && qr >= r) return maxv[rt]; int mid = (l + r) >> 1, ret = -1e9; pushdown(rt); if(ql <= mid) ret = max(ret, query(lson, ql, qr)); if(qr > mid) ret = max(ret, query(rson, ql, qr)); pushup(rt); return ret;}int ask(int x, int y) { int ret = -1e9; while(top[x] != top[y]) { if(dep[top[x]] < dep[top[y]]) swap(x, y); ret = max(ret, query(1, 1, idcnt, tid[top[x]], tid[x])); x = fa[top[x]]; } if(dep[x] > dep[y]) swap(x, y); if(x != y) ret = max(ret, query(1, 1, idcnt, tid[x] + 1, tid[y])); return ret;}void Negate(int x, int y) { while(top[x] != top[y]) { if(dep[top[x]] < dep[top[y]]) swap(x, y); update(1, 1, idcnt, tid[top[x]], tid[x]); x = fa[top[x]]; } if(dep[x] > dep[y]) swap(x, y); if(x != y) update(1, 1, idcnt, tid[x] + 1, tid[y]);}void debug() { for(int i = 1; i <= n; i++) { printf("for %d top is %d tid is %d val is %d\n", i, top[i], tid[i], sval[tid[i]]); } printf("the maxv: "); for(int i = 1; i <= n << 2; i++) printf("%d ", maxv[i]); puts("");}int main() { int T; scanf("%d", &T); while(T--) { init(); scanf("%d", &n); for(int i = 1; i < n; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); adde(a, b, c, i); adde(b, a, c, i); } dfs1(1, 1, 1); dfs2(1, 1); for(int i = 0; i < ecnt; i++) { if(dep[u[i]] > dep[v[i]]) { sval[tid[u[i]]] = w[i]; rnum[eid[i]] = tid[u[i]]; } else { sval[tid[v[i]]] = w[i]; rnum[eid[i]] = tid[v[i]]; } } build(1, 1, idcnt); //debug(); char cmd[16]; int x, y; while(scanf("%s", cmd)) { if(strcmp(cmd, "DONE") == 0) break; scanf("%d%d", &x, &y); if(cmd[0] == ‘Q‘) { printf("%d\n", ask(x, y)); } else if(cmd[0] == ‘C‘) { change(1, 1, idcnt, rnum[x], y); } else { Negate(x, y); } } } return 0;}
POJ 3237 Tree 树链剖分
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。