hdu 1540 Tunnel Warfare 线段树 / set
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题目链接
题意
一排数字
法一:线段树
思路
维护很常规,维护区间内 左起连续的个数,右起连续的个数,最大连续的长度。
询问时比较独特,需要额外的判断:如果询问的
Code
#include <bits/stdc++.h>#define lson (rt << 1)#define rson (rt << 1 | 1)#define maxn 50010using namespace std;stack<int> s;bool brok[maxn];struct node { int l, r, len, ml, mr, mm; }tr[maxn * 4];void push_up(int rt) { tr[rt].ml = tr[lson].ml; if (tr[lson].ml == tr[lson].len) tr[rt].ml += tr[rson].ml; tr[rt].mr = tr[rson].mr; if (tr[rson].mr == tr[rson].len) tr[rt].mr += tr[lson].mr; tr[rt].mm = max(tr[lson].mm, max(tr[rson].mm, tr[lson].mr + tr[rson].ml));}void build(int rt, int l, int r) { tr[rt].l = l, tr[rt].r = r, tr[rt].len = r - l + 1; tr[rt].ml = tr[rt].mr = tr[rt].mm = tr[rt].len; if (l == r) return; int mid = l + r >>1; build(lson, l, mid); build(rson, mid + 1, r);}void modify(int rt, int x, int w) { if (tr[rt].l == tr[rt].r) { tr[rt].ml = tr[rt].mr = tr[rt].mm = w; return; } int mid = tr[rt].l + tr[rt].r >> 1; if (x <= mid) modify(lson, x, w); else modify(rson, x, w); push_up(rt);}int query(int rt, int x) { if (tr[rt].l == tr[rt].r || tr[rt].mm == tr[rt].len || tr[rt].mm == 0) return tr[rt].mm; int mid = tr[rt].l + tr[rt].r >> 1; if (x <= mid) { if (x > tr[lson].r - tr[lson].mr) return query(lson, x) + query(rson, mid + 1); else return query(lson, x); } else { if (x < tr[rson].l + tr[rson].ml) return query(rson, x) + query(lson, mid); else return query(rson, x); }}int n, m;void work() { while (!s.empty()) s.pop(); memset(brok, 0, sizeof brok); build(1, 1, n); while (m--) { char c; int x; scanf("\n%c", &c); if (c == 'D') { scanf("%d", &x); if (!brok[x]) { brok[x] = true; modify(1, x, 0); } s.push(x); } else if (c == 'R') { if (s.empty()) continue; int x = s.top(); if (brok[x]) { brok[x] = false; modify(1, x, 1); } s.pop(); } else { scanf("%d", &x); printf("%d\n", query(1, x)); } }}int main() { while (scanf("%d%d", &n, &m) != EOF) work(); return 0;}
法二:set
参考
HDU1540 Tunnel Warfare(线段树+set) ——winjourn
思路
用一个
(太方便好写了(泪目
Code
#include <bits/stdc++.h>#define maxn 50010using namespace std;stack<int> s;bool brok[maxn];set<int> st;int n, m;void work() { while (!s.empty()) s.pop(); st.clear(); memset(brok, 0, sizeof brok); while (m--) { char c; int x; scanf("\n%c", &c); if (c == 'D') { scanf("%d", &x); if (!brok[x]) { brok[x] = true; st.insert(x); } s.push(x); } else if (c == 'R') { if (s.empty()) continue; int x = s.top(); if (brok[x]) { brok[x] = false; st.erase(x); } s.pop(); } else { scanf("%d", &x); int ans; if (st.empty()) ans = n; else { auto p = lower_bound(st.begin(), st.end(), x); if (p == st.end()) ans = n - *(--p); else if (*p == x) ans = 0; else { if (p == st.begin()) ans = *p - 1; else ans = *p - *(--p) - 1; } } printf("%d\n", ans); } }}int main() { while (scanf("%d%d", &n, &m) != EOF) work(); return 0;}
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