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Codeforces Round #254 (Div. 2)
Codeforces Round #254 (Div. 2)
题目链接
A题:给定一个棋盘,放B,W不能相邻,输出摆法
思路:模拟国际象棋,B放在白格,A放在黑格即可
B题:给定一些化学物品,给定哪些可以反应,现在一一加入试管,如果试管之前有加过可以反应的,危险度乘2,初始危险度为1,求最小危险度
思路:用并查集,找出有多少个集合,这些先加进去保证不会反应,那么剩下的一个个加进去都乘2即可
C题:求一个最小密度连通子图,子图必须为一个集合,并且两点存在的话,他们的边必须存在
思路:推一推公式发现是选两个点,密度最大的最优,再往后加一个点都会导致密度变小
D题:根据题目给的函数生成A和B数组,求出长度1-n的子数组中的max(ai*bn-i)
思路:如果1比较多的情况,那么之后从后往前找30个左右很快就能找到答案,如果1比较少的情况,那么直接从前往后,在1的情况上找,这样做的合理性在于,对于b数组,基本1的分布情况是30多个左右就会有1个1,所以总体复杂度并不会太高
E题:一个区间,两种操作,可以染上某种颜色,使每个位置增量增加染上和现有颜色之差绝对值,还有一种操作是询问某个区间的总增量
明显的线段树,区间询问区间查询,利用延迟操作求解
代码:
A:
#include <stdio.h> #include <string.h> int n, m; char g[105][105]; int main() { scanf("%d%d", &n, &m); for (int i = 0; i < n; i++) scanf("%s", g[i]); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (g[i][j] == '-') continue; if ((i + j) % 2 == 0) g[i][j] = 'B'; else g[i][j] = 'W'; } } for (int i = 0; i < n; i++) printf("%s\n", g[i]); return 0; }
B:
#include <stdio.h> #include <string.h> int n, m; int parent[55]; int find(int x) { if (x == parent[x]) return x; return parent[x] = find(parent[x]); } int main() { scanf("%d%d", &n, &m); int ans = n; int x, y; for (int i = 1; i <= n; i++) parent[i] = i; while (m--) { scanf("%d%d", &x, &y); int pa = find(x); int pb = find(y); if (pa != pb) { parent[pa] = pb; n--; } } long long out = 1; for (int i = 0; i < ans - n; i++) out *= 2; printf("%lld\n", out); return 0; }
C:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int N = 505; int n, m; double v[N]; int main() { scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) scanf("%lf", &v[i]); int a, b; double w; double ans = 0; while (m--) { scanf("%d%d%lf", &a, &b, &w); ans = max(ans, (v[a] + v[b]) / w); } printf("%.15lf\n", ans); return 0; }
D:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 100005; int a[N], b[N], n, d, x, to[N], q[N]; int getnext() { x = (x * 37LL + 10007) % 1000000007; return x; } void initAB() { for (int i = 0; i < n; i++) a[i] = i + 1; for (int i = 0; i < n; i++) swap(a[i], a[getnext() % (i + 1)]); for (int i = 0; i < n; i++) { if (i < d) b[i] = 1; else b[i] = 0; } for (int i = 0; i < n; i++) swap(b[i], b[getnext() % (i + 1)]); } int main() { scanf("%d%d%d", &n, &d, &x); initAB(); for (int i = 0; i < n; i++) { to[a[i]] = i; if (b[i]) q[++q[0]] = i; } int s = 30; for (int i = 0; i < n; i++) { int j; for (j = n; j >= n - s; j--) { if (j > 0 && i >= to[j] && b[i - to[j]]) { printf("%d\n", j); break; } } if (j < n - s) { int ans = 0; for (int j = 1; j <= q[0] && q[j] <= i; j++) ans = max(ans, a[i - q[j]]); printf("%d\n", ans); } } return 0; }
E:
#include <stdio.h> #include <string.h> #include <algorithm> using namespace std; const int N = 100005; typedef long long ll; inline int lson(int x) {return x<<1;} inline int rson(int x) {return x<<1|1;} struct Node { int l, r; ll sum, col, add; int size() {return r - l + 1;} int mid() {return (r + l)>>1;} } node[N * 4]; void pushup(int x) { node[x].col = (node[lson(x)].col == node[rson(x)].col?node[lson(x)].col : 0); node[x].sum = node[lson(x)].sum + node[rson(x)].sum; } void pushdown(int x) { if (node[x].add) { node[lson(x)].add += node[x].add; node[rson(x)].add += node[x].add; node[lson(x)].sum += node[x].add * node[lson(x)].size(); node[rson(x)].sum += node[x].add * node[rson(x)].size(); node[lson(x)].col = node[rson(x)].col = node[x].col; node[x].col = node[x].add = 0; } } void build(int x, int l, int r) { node[x].l = l; node[x].r = r; node[x].sum = node[x].col = node[x].add = 0; if (l == r) { node[x].col = l; return; } build(lson(x), l, node[x].mid()); build(rson(x), node[x].mid() + 1, r); pushup(x); } void update(int x, int l, int r, ll col) { if (node[x].l >= l && node[x].r <= r && node[x].col) { node[x].add += abs(node[x].col - col); node[x].sum += abs(node[x].col - col) * node[x].size(); node[x].col = col; return; } pushdown(x); if (l <= node[x].mid()) update(lson(x), l, r, col); if (r > node[x].mid()) update(rson(x), l, r, col); pushup(x); } long long query(int x, int l, int r) { if (node[x].l >= l && node[x].r <= r) return node[x].sum; pushdown(x); long long ans = 0; if (l <= node[x].mid()) ans += query(lson(x), l, r); if (r > node[x].mid()) ans += query(rson(x), l, r); return ans; } int n, m; int main() { scanf("%d%d", &n, &m); build(1, 1, n); int q, l, r; ll x; while (m--) { scanf("%d", &q); if (q == 1) { scanf("%d%d%lld", &l, &r, &x); update(1, l, r, x); } else { scanf("%d%d", &l, &r); printf("%lld\n", query(1, l, r)); } } return 0; }
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