CodeForces AIM Tech Round 3 (Div. 1) 题解(CF708A,CF708B,CF708C,CF708D,CF708E)
2017-02-21 08:50
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A. Letters Cyclic Shift(CodeForces 708A)
题解:直接找到第一个不是a的位置i和i后面第一个为a的位置j(如果没有,j=len(s)),那么翻转的肯定就是[i,j]这一段。如果这个字符串都是a,那么就翻转最后一个字符。#include <cstring> #include <cstdio> #include <algorithm> using namespace std; char s[100005]; int main() { scanf("%s", s + 1); int n = strlen(s + 1); bool flag = 0; for (int i = 1; i <= n; i ++) { if (!flag && s[i] != 'a') flag = 1; if (flag && s[i] == 'a') { for (int j = i; j <= n; j ++) printf("%c", s[j]); break; } if (!flag && i != n) printf("%c", s[i]); else printf("%c", (s[i] - 'a' - 1 + 26) % 26 + 'a'); } }
B. Recover the String(CodeForces 708B)
题解:由于知道了A[0][0],假设0的数量是n0,那么A[0][0]肯定等于n0*(n0-1)/2,这样就可以算出n0,同理算出n1,同时判断A[0][0]和A[1][1]是否合法,同时也可以判断A[0][0]+A[0][1]+A[1][0]+A[1][1]是否等于(n0+n1)*(n0+n1-1)/2,即总数是否合法。最后根据A[0][1]的信息从前往后判断0是否可以填到第i位,满足0肯定肯定合法,同时1的限制肯定也合法了。还有一种特殊情况是当A[0][0]=A[0][1]=A[1][0]=0时n0等于0,1的情况同理。最后特判一下A[0][0]+A[0][1]+A[1][0]+A[1][1]=0的情况就可以了。#include <cstring> #include <cstdio> #include <algorithm> #include <cmath> using namespace std; typedef long long LL; LL a, b, c, d; char s[1000005]; int main() { scanf("%lld%lld%lld%lld", &a, &b, &c, &d); if (a == 0 && b == 0 && c == 0 && d == 0) {printf("1\n"); return 0;} LL num0 = sqrt(a * 2) + 1, num1 = sqrt(d * 2) + 1; if (a == 0 && b == 0 && c == 0) num0 = 0; if (b == 0 && c == 0 && d == 0) num1 = 0; if (num0 * (num0 - 1) / 2 != a || num1 * (num1 - 1) / 2 != d) {printf("Impossible\n"); return 0;} if (b + c != num0 * num1) {printf("Impossible\n"); return 0;} LL all = num0 + num1; LL e = all * (all - 1) / 2; if (all > 1000000 || e != a + b + c + d) {printf("Impossible\n"); return 0;} LL x = num0, y = num1, bb = b, cc = 0; for (int i = 1; i <= all; i ++) { if (y <= bb) s[i] = '0', bb -= y; else s[i] = '1', y --; } printf("%s", s + 1); }
C. Centroids(CodeForces 708C)
题解:首先加入一个点能成为重心,那么就要存在节点数不大于n/2的子树。由于最多只有一个,我们只需要维护每个节点向下能切下来小于等于n/2的最大子树和向上能切下来小于等于n/2的最大子树。分别用一个树形dp来求出,最后判断每个节点最大的子树是否合法就可以了。#include <cstring> #include <cstdio> #include <algorithm> #include <set> using namespace std; const int MAXN = 4e5 + 5; int n, size[MAXN], hev[MAXN], down[MAXN], up[MAXN]; int tot, Last[MAXN], Go[MAXN * 2], Next[MAXN * 2]; void link(int u, int v) { Next[++ tot] = Last[u], Last[u] = tot, Go[tot] = v; } void dfs1(int now, int pre) { size[now] = 1; for (int p = Last[now]; p; p = Next[p]) { int v = Go[p]; if (v == pre) continue; dfs1(v, now); size[now] += size[v]; down[now] = max(down[now], (size[v] <= n / 2) ? size[v] : down[v]); hev[now] = max(hev[now], size[v]); } } void dfs2(int now, int pre) { multiset<int> d; for (int p = Last[now]; p; p = Next[p]) { int v = Go[p]; if (v != pre) d.insert((size[v] <= n / 2) ? size[v] : down[v]); } for (int p = Last[now]; p; p = Next[p]) { int v = Go[p]; if (v == pre) continue; if (n - size[v] <= n / 2) up[v] = max(up[v], n - size[v]); else { up[v] = max(up[v], up[now]); d.erase(d.find(size[v] <= n / 2 ? size[v] : down[v])); if (!d.empty()) up[v] = max(up[v], *d.rbegin()); d.insert((size[v] <= n / 2) ? size[v] : down[v]); } dfs2(v, now); } } int main() { scanf("%d", &n); for (int i = 1; i < n; i ++) { int u, v; scanf("%d%d", &u, &v); link(u, v), link(v, u); } dfs1(1, 0); dfs2(1, 0); for (int i = 1; i <= n; i ++) { int ans = 1; if (n - size[i] > n / 2) ans = (n - size[i] - up[i] <= n / 2); if (hev[i] > n / 2) ans = (hev[i] - down[i] <= n / 2); printf("%d ", ans); } }
D. Incorrect Flow(CodeForces 708D)
题解:设(u,v,f,w)表示从u到v有一条流量为f,费用为w的边。假设我们先连(n,1,Inf,0),那么最终保证流量平衡就可以了。对于c>=f的边很好连,直接(u,v,c-f,1),(u,v,Inf,2),(v,u,f,1),对于c#include <cstring> #include <cstdio> #include <algorithm> using namespace std; const int Inf = 1e9; const int MAXN = 105, MAXM = 505; int n, m, S, T, ans, deg[MAXN], dis[MAXN], d[MAXN]; int tot, Last[MAXN], Go[MAXM * 2], Next[MAXM * 2], Len[MAXM * 2], Cost[MAXM * 2]; bool vis[MAXN], flag[MAXN]; void open(int u, int v, int f, int w) { Next[++ tot] = Last[u], Last[u] = tot, Go[tot] = v, Len[tot] = f, Cost[tot] = w; } void link(int u, int v, int f, int w) { open(u, v, f, w), open(v, u, 0, -w); } int dfs(int now, int flow) { if (now == T) { ans += dis[T] * flow; return flow; } int use = 0; vis[now] = 1; for (int p = Last[now]; p; p = Next[p]) { int v = Go[p]; if (Len[p] && !vis[v] && dis[v] == dis[now] + Cost[p]) { int t = dfs(v, min(flow - use, Len[p])); Len[p] -= t, Len[p ^ 1] += t, use += t; if (use == flow) return use; } } return use; } int solve() { while (1) { memset(vis, 0, sizeof vis), memset(dis, 60, sizeof dis); int l = 0, r = 1; d[1] = S, dis[S] = 0, flag[S] = 1; while (l != r) { l = (l + 1) % MAXN; int now = d[l]; for (int p = Last[now]; p; p = Next[p]) { int v = Go[p]; if (Len[p] && dis[now] + Cost[p] < dis[v]) { dis[v] = dis[now] + Cost[p]; if (!flag[v]) { flag[v] = 1; d[r = (r + 1) % MAXN] = v; } } } flag[now] = 0; } if (dis[T] == dis[MAXN - 1]) return ans; dfs(S, Inf); } } int main() { scanf("%d%d", &n, &m); S = 0, T = n + 1, tot = 1; link(n, 1, Inf, 0); for (int i = 1; i <= m; i ++) { int u, v, c, f; scanf("%d%d%d%d", &u, &v, &c, &f); deg[u] -= f, deg[v] += f; if (c >= f) { link(u, v, c - f, 1); link(u, v, Inf, 2); link(v, u, f, 1); } else { ans += f - c; link(u, v, Inf, 2); link(v, u, c, 1); link(v, u, f - c, 0); } } for (int i = 1; i <= n; i ++) if (deg[i] < 0) link(i, T, -deg[i], 0); else link(S, i, deg[i], 0); solve(); printf("%d\n", ans); }
E. Student’s Camp(CodeForces 708E)
施工中……相关文章推荐
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