2022 CCPC 华为云计算挑战赛 机器人

题目链接

其实是补2023CCPC秦皇岛热身赛C

主要思路跟IOI2021分糖果是一样的，区别就是这里不是对总的区间二分，而是指定区间

所以先做一次区间询问把对应的log个线段树区间拿出来，然后就是二分一样的思路，不过是在序列上，所以要先逆序找到第一个不满足条件的线段树区间，然后进到它对应的子树里二分即可

常数巨大

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#include <iostream> #include <algorithm> #include <vector> #define ls (x << 1) #define rs (x << 1 | 1)   using namespace std;   typedef long long LL; const int MAX_N = 5e5 + 5, MIN_V = -2e9, MAX_V = 2e9;   int opt[2][MAX_N], sig[2][MAX_N]; vector<tuple<int,int,int>> q[2]; struct NODE{     LL lzy;     int maxv, minv;     NODE(LL cl = 0, int cx = 0, int cn = 0) : lzy{cl}, maxv{cx}, minv{cn} {} }; struct SGT{     NODE t[MAX_N << 2];     void pushup(int x) {         t[x].maxv = max(t[ls].maxv, t[rs].maxv);         t[x].minv = min(t[ls].minv, t[rs].minv);     }     void Merge(NODE &x, NODE l, NODE r) {         x.maxv = max(l.maxv, r.maxv);         x.minv = min(l.minv, r.minv);     }     void pushdown(int x) {         LL v = t[x].lzy; t[x].lzy = 0;         t[ls].lzy += v; t[ls].minv += v; t[ls].maxv += v;         t[rs].lzy += v; t[rs].minv += v; t[rs].maxv += v;     }     void update(int L, int R, int l, int r, int x, int v) {         if (L <= l && r <= R) {             t[x].lzy += v; t[x].minv += v; t[x].maxv += v;             return;         }         pushdown(x); int mid = l + r >> 1;         if (L <= mid) update(L, R, l, mid, ls, v);         if (mid < R) update(L, R, mid + 1, r, rs, v);         pushup(x);     }     int getval(int D, int l, int r, int x) {         if (l == r) return t[x].minv;         pushdown(x); int mid = l + r >> 1;         if (D <= mid) return getval(D, l, mid, ls);         else return getval(D, mid + 1, r, rs);     }     void ask(int L, int R, int l, int r, int x, int op) {         if (L <= l && r <= R) {             q[op].push_back({x, l, r}); return;         }         pushdown(x); int mid = l + r >> 1;         if (L <= mid) ask(L, R, l, mid, ls, op);         if (mid < R) ask(L, R, mid + 1, r, rs, op);     }     int getans(int l, int r, int n, int m, int op) {         NODE suf(0, MIN_V, MAX_V), nxt(0, MIN_V, MAX_V);         int x = 0, lef = 0, rig = 0;         for (int i = q[op].size() - 1; i >= 0; --i) {             tie(x, lef, rig) = q[op][i];             Merge(nxt, t[x], suf);             if (nxt.maxv - nxt.minv > n) {                 return getval(r, 0, m, 1) + query(lef, rig, x, n, suf);             }             suf = nxt;         }         return getval(r, 0, m, 1) - suf.minv;     }     int query(int l, int r, int x, int top, NODE suf) {         if (l == r) {             NODE cur = NODE(0, MIN_V, MAX_V); Merge(cur, t[x], suf);             if (cur.minv == t[x].minv) return top - cur.maxv;             else return -cur.minv;         }         pushdown(x); int mid = l + r >> 1;         NODE nxt = NODE(0, MIN_V, MAX_V); Merge(nxt, t[rs], suf);         if (nxt.maxv - nxt.minv > top) return query(mid + 1, r, rs, top, suf);         else return query(l, mid, ls, top, nxt);     }     void build(int l, int r, int x, int op) {         t[x].lzy = t[x].maxv = t[x].minv = 0;         if (l == r) {             t[x].maxv = t[x].minv = sig[op][l];             return;         }         int mid = l + r >> 1;         build(l, mid, ls, op);         build(mid + 1, r, rs, op);         pushup(x);     } }t[2]; char s[MAX_N]; int stk[30]; int rd() {     int x=  0, c = getchar();     while (!isdigit(c)) c = getchar();     while (isdigit(c)) {         x = (x << 1) + (x << 3) + (c ^ 48);         c = getchar();     }     return x; } void rds() {     int c = getchar(), ptr = 0;     while (!isupper(c)) c = getchar();     while (isupper(c)) {         s[++ptr] = c;         c = getchar();     } } void output(int x) {     int top = 0;     do {         stk[top++] = x % 10; x /= 10;     } while (x);     while (top) {         putchar('0' + stk[--top]);     } } char rdc() {     int c = getchar();     while (!isupper(c)) c = getchar();     return c; } void solve() {     int n, m, k; n = rd(); m = rd(); k = rd();     rds();     for (int i = 1; i <= m; ++i) {         opt[0][i] = opt[1][i] = 0;         if (s[i] == 'U' || s[i] == 'D') {             opt[0][i] = (s[i] == 'U' ? -1 : 1);         }         if (s[i] == 'L' || s[i] == 'R') {             opt[1][i] = (s[i] == 'L' ? -1 : 1);         }         sig[0][i] = sig[0][i - 1] + opt[0][i];         sig[1][i] = sig[1][i - 1] + opt[1][i];     }     t[0].build(0, m, 1, 0); t[1].build(0, m, 1, 1);     int op, x[2], l, r, pos; char c;     while (k--) {         op = rd();         if (op == 1) {             x[0] = rd() - 1; x[1] = rd() - 1; l = rd(); r = rd();             for (int i = 0; i < 2; ++i) {                 if (x[i]) t[i].update(l, r, 0, m, 1, x[i]);                 q[i].clear(); t[i].ask(l - 1, r, 0, m, 1, i);                 output(t[i].getans(l - 1, r, n - 1, m, i) + 1);                 if (i == 0) putchar(' ');                 if (x[i]) t[i].update(l, r, 0, m, 1, -x[i]);             } putchar('\n');         } else {             pos = rd(); c = rdc();             int cur = (s[pos] == 'U' || s[pos] == 'D') ? 0 : 1;             if (opt[cur][pos]) t[cur].update(pos, m, 0, m, 1, -opt[cur][pos]);             opt[cur][pos] = 0;             if (c == 'U' || c == 'D') {                 opt[0][pos] = (c == 'U' ? -1 : 1);                 t[0].update(pos, m, 0, m, 1, opt[0][pos]);             } else {                 opt[1][pos] = (c == 'L' ? -1 : 1);                 t[1].update(pos, m, 0, m, 1, opt[1][pos]);             }             s[pos] = c;         }     } } int main() {     int T = 1; T = rd();     while (T--) {         solve();     }     return 0; }