A Game with Colored Balls
- 题意:
给一个长度为n的字符串,每次删除字母同样切连续的串,假设有多个,删除最左边的、最长的串。每次删除输出串的字母,每一个字母的下标(1-n)
N (1 ≤ N ≤ 106),串仅仅包含red (‘R’), green (‘G’) or blue (‘B’)
- 分析:
题目比較麻烦,分析一下须要的任务:
1、每次找到最长串——优先队列
2、删除最长串后,须要合并两側的串(假设字母同样)——加Hash的链表,set(超时)
3、每次删除须要知道这一个区间的下标都是谁——加Hash的链表,set(超时)
C++通过,G++超时……
const int MAXN = 1100000; struct Node { int pos, len; char val; Node *nxt, *pre; Node (int p = 0, int n = 0, char v = 0) : pos(p), len(n), val(v) {} bool operator< (const Node& rhs) const { return pos < rhs.pos; } void erase() { pre->nxt = nxt; nxt->pre = pre; } } nd[MAXN], pt, fst, lst; int tot; struct HeapNode { int val, pos; HeapNode (int v = 0, int p = 0) : val(v), pos(p) {} bool operator< (const HeapNode& rhs) const { if (val != rhs.val) return val < rhs.val; return pos > rhs.pos; } } vt; priority_queue<HeapNode> q; char ipt[MAXN]; bool vis[MAXN]; Node* to[MAXN], *pit, *pl, *pr; int nxt[MAXN], pre[MAXN]; void init(int n) { REP(i, n) { nxt[i] = i + 1; if (i) pre[i] = i - 1; } } void erase(int l, int r) { int p = pre[l], n = nxt[r]; nxt[p] = n; pre[n] = p; } int main() { while (~RS(ipt)) { fst.val = -1; fst.nxt = &lst; fst.pre = NULL; lst.val = -2; lst.pre = &fst; lst.nxt = NULL; CLR(vis, false); while (!q.empty()) q.pop(); tot = 0; int len = strlen(ipt); init(len + 2); nd[tot++] = Node(1, 1, ipt[0]); FF(i, 1, len) { if (ipt[i] == nd[tot - 1].val) nd[tot - 1].len++; else { nd[tot].pos = i + 1; nd[tot].len = 1; nd[tot].val = ipt[i]; tot++; } } fst.nxt = &nd[0]; nd[0].pre = &fst; lst.pre = &nd[tot - 1]; nd[tot - 1].nxt = &lst; REP(i, tot) { if (i != 0) nd[i].pre = &nd[i - 1]; if (i != tot - 1) nd[i].nxt = &nd[i + 1]; to[nd[i].pos] = &nd[i]; q.push(HeapNode(nd[i].len, nd[i].pos)); } while (!q.empty()) { vt = q.top(); q.pop(); if (vt.val == 1) break; if (vis[vt.pos]) continue; pt.pos = vt.pos; pit = to[vt.pos]; int idx = vt.pos; printf("%c", ipt[vt.pos - 1]); REP(i, vt.val) { printf(" %d", idx); erase(idx, idx); idx = nxt[pre[idx]]; } puts(""); pl = pit->pre; pr = pit->nxt; if (pl->val == pr->val) { pl->len += pr->len; vis[pr->pos] = true; pr->erase(); q.push(HeapNode(pl->len, pl->pos)); } vis[vt.pos] = true; pit->erase(); } } return 0; }