FOJ月赛 2015年3月
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A:把第一个字符串预处理一下,记录字母,和每个字母的出现次数,然后在第一个字符串枚举开始的位置,然后和第二个字符串进行比较,对于第一个字母,要特殊处理,因为可能是从该字母的第x个位置开始算,这样时间复杂度为O(n^2)
B:简单的树状数组应用,每次取出当前第x + 1大的数字即可
C:第一问:答案为(度数为1的结点+1)/ 2,第二问,其实也就是贪心,每次都尽量找最长链,看这个子树有x条链,答案就加上(x + 1) / 2
D:先BFS预处理出宝藏(起点也当成一个宝藏)之间的最短路,然后进行状压DP,dp[i][j]表示拥有宝藏的状态i和当前在j宝藏的最小代价,最后在枚举一下记录答案即可,有个坑就是一开始起点可以是墙,答案是-1
E:线段树处理面积并的模板题。
F:这题数据小,记忆化广搜一下即可,d[i][j][2]表示,左岸i只羊j只狼,0表示当前在左岸,1表示当前在右岸,然后进行状态转移,注意细节比较多,比如如果羊数量是0,那么狼有几只都没问题
G:这题数据大了,不能搜了,就直接贪心,对于n <= 3还是去搜就行了,这时候m大于6的就没答案了,所以可以搜,然后对于n > 3时,在羊和狼都比较多的时候,可以每次运n只过去(狼羊对半开),然后运一对羊狼回来,直到数量少了,进行贪心,这里贪心分为3类和一种特殊情况(这个一开始没考虑到,靠前一个程序打表发现的- -),3种情况就是n == m时候,要5次,m * 2 <= n时候要1次,在这之间的要3次。特殊情况为:举个例子,羊和狼都有6只,船能运5只,那么第一次可以运3只狼2只羊,然后运1只狼回来,这时候羊只有4只了,就可以尽量多的把羊都运过去,羊变成0了,这样对左岸就不在有影响了,也就是说,船在奇数情况下,能充分利用这多出来的一个位置,答案会更优,这种情况下,直接在多5次就可以了(而不是先常规运一次在判断,这样可能要7次)。
代码:
A:
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1005;char a[N], b[N];int cnt[N];void build() { int n = strlen(a); int an = 0; for (int i = 0 ; i < n; i++) { if (a[i] >= 'a' && a[i] <= 'z') { a[an] = a[i]; cnt[an] = 1; if (a[i + 1] == '[') { int num = 0; for (int j = i + 2; j < n; j++) { if (a[j] == ']') { cnt[an] = num; i = j; break; } num = num * 10 + a[j] - '0'; } } an++; } } a[an] = '\0';}bool judge() { int n = strlen(a); int m = strlen(b); for (int i = 0; i < n; i++) { int x; if (b[0] != a[i]) continue; for (x = 0; x < cnt[i] && x < m; x++) { if (b[x] != a[i]) break; } for (int u = i + 1; u < n && x < m; u++) { int flag = 0; for (int j = 0; j < cnt[u] && x < m; j++) { if (b[x] != a[u]) { flag = 1; break; } x++; } if (flag) break; } if (x == m) return true; } return false;}int main() { while (~scanf("%s%s", a, b)) { build(); if (judge()) printf("True\n"); else printf("False\n"); } return 0;}
B:
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;inline int lowbit(int x) { return x & (-x);}const int N = 1005;int n;int bit[N];void add(int x, int v) { while (x <= n) { bit[x] += v; x += lowbit(x); }}int get(int x) { int ans = 0; while (x) { ans += bit[x]; x -= lowbit(x); } return ans;}int find(int x) { int l = 0, r = n; while (l < r) { int mid = (l + r) / 2; if (get(mid) >= x) r = mid; else l = mid + 1; } return r;}int main() { while (~scanf("%d", &n)) { memset(bit, 0, sizeof(bit)); for (int i = 1; i <= n; i++) add(i, 1); int x; int bo = 0; for (int i = 0; i < n; i++) { scanf("%d", &x); int tmp = find(x + 1); if (bo) printf(" "); else bo = 1; printf("%d", tmp); add(tmp, -1); } printf("\n"); } return 0;}
C:
#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int N = 100005;int n, f, du[N], ans;vector<int> g[N];void dfs(int u, int p, int flag) { int sum = 0; for (int i = 0; i < g[u].size(); i++) { int v = g[u][i]; if (v == p) continue; sum++; dfs(v, u, 1); } ans += (sum - flag + 1) / 2;}int t;int main() { scanf("%d", &t); while (t--) { scanf("%d", &n); ans = 0; for (int i = 0; i <= n - 1; i++) { g[i].clear(); du[i] = 0; } for (int i = 1; i <= n - 1; i++) { scanf("%d", &f); du[f]++; du[i]++; g[f].push_back(i); g[i].push_back(f); } int cnt = 0; for (int i = 0; i <= n - 1; i++) if (du[i] == 1) cnt++; if (n == 1) cnt = 0; else cnt = (cnt + 1) / 2; dfs(0, -1, 0); printf("%d %d\n", cnt, ans); } return 0;}
D:
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int N = 105;const int INF = 0x3f3f3f3f;const int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};int n, m, G[N][N], g[15][15];int x[15], y[15], cnt;struct State { int x, y; State() {} State(int x, int y) { this->x = x; this->y = y; }};int dis[105][105], vis[105][105];void build(int u) { queue<State> Q; Q.push(State(x[u], y[u])); memset(vis, 0, sizeof(vis)); for (int i = 0; i < n; i++) for(int j = 0; j < m; j++) dis[i][j] = INF; dis[x[u]][y[u]] = 0; vis[x[u]][y[u]] = 1; while (!Q.empty()) { State uu = Q.front(); Q.pop(); vis[uu.x][uu.y] = 0; for (int i = 0; i < 4; i++) { State v; v.x = uu.x + d[i][0]; v.y = uu.y + d[i][1]; if (v.x < 0 || v.x > n || v.y <0 || v.y> m || G[v.x][v.y] < 0) continue; if (dis[v.x][v.y] > dis[uu.x][uu.y] + 1) { dis[v.x][v.y] = dis[uu.x][uu.y] + 1; if (!vis[v.x][v.y]) { vis[v.x][v.y] = 1; Q.push(v); } } } } for (int i = 1; i <= cnt; i++) g[u][i] = dis[x[i]][y[i]];}int dp[2100][15];int main() { while (~scanf("%d%d", &n, &m)) { cnt = 0; int cao = 0; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) { scanf("%d", &G[i][j]); if (i == 0 && j == 0) { if (G[i][j] < 0) { cao = 1; } } if ((i == 0 && j == 0) || G[i][j] > 0) { ++cnt; x[cnt] = i; y[cnt] = j; G[i][j] = cnt; } } if (cao == 1) { printf("-1\n"); continue; } for (int i = 1; i <= cnt; i++) { for (int j = 1; j <= cnt; j++) g[i][j] = INF; } for (int i = 1; i <= cnt; i++) build(i); dp[1][0] = 0; for (int i = 1; i < (1<<cnt); i++) { for (int j = 0; j < cnt; j++) { if (i == 1 && j == 0)continue; dp[i][j] = INF; if (i&(1<<j)) { for (int k = 0; k < cnt; k++) { if (j == k) continue; if (g[j + 1][k + 1] == INF) continue; if (i&(1<<k)) dp[i][j] = min(dp[i][j], dp[i^(1<<j)][k] + g[k + 1][j + 1]); } } } } int ans = INF; for (int i = 1; i <= cnt; i++) { ans = min(ans, dp[(1<<cnt) - 1][i - 1] + g[1][i]); } if (ans == INF) ans = -1; printf("%d\n", ans); } return 0;}
E:
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int N = 100005;typedef long long ll;int t, n;struct Line { int l, r, y, flag; Line() {} Line(int l, int r, int y, int flag) { this->l = l; this->r = r; this->y = y; this->flag = flag; }} line[N * 2];bool cmp(Line a, Line b) { return a.y < b.y;}int Hash[N * 2], ln, hn;int find(int x) { return lower_bound(Hash, Hash + hn, x) - Hash;}#define lson(x) ((x<<1)+1)#define rson(x) ((x<<1)+2)struct Node { int l, r, cover, len[3]; void init(int l, int r) { this->l = l; this->r = r; cover = 0; memset(len, 0, sizeof(len)); }} node[N * 8];void build(int l, int r, int x = 0) { node[x].init(l, r); node[x].len[0] = Hash[r + 1] - Hash[l]; if (l == r) return; int mid = (l + r) / 2; build(l, mid, lson(x)); build(mid + 1, r, rson(x));}void pushup(int x) { memset(node[x].len, 0, sizeof(node[x].len)); if (node[x].l == node[x].r) { node[x].len[min(2, node[x].cover)] = Hash[node[x].r + 1] - Hash[node[x].l]; return; } for (int i = 0; i <= 2; i++) node[x].len[min(2, node[x].cover + i)] += (node[lson(x)].len[i] + node[rson(x)].len[i]);}void add(int l, int r, int v, int x = 0) { if (node[x].l >= l && node[x].r <= r) { node[x].cover += v; pushup(x); return; } int mid = (node[x].l + node[x].r) / 2; if (l <= mid) add(l, r, v, lson(x)); if (r > mid) add(l, r, v, rson(x)); pushup(x);}ll solve() { ln = hn = 0; int x1, y1, x2, y2; for (int i = 0; i < n; i++) { scanf("%d%d%d%d", &x1, &y1, &x2, &y2); line[ln++] = Line(x1, x2, y1, 1); line[ln++] = Line(x1, x2, y2, -1); Hash[hn++] = x1; Hash[hn++] = x2; } sort(line, line + ln, cmp); sort(Hash, Hash + hn); hn = unique(Hash, Hash + hn) - Hash; ll ans = 0; build(0, hn - 2); for (int i = 0; i < ln; i++) { if (i) ans += (ll)node[0].len[1] * (ll)(line[i].y - line[i - 1].y); add(find(line[i].l), find(line[i].r) - 1, line[i].flag); } return ans;}int main() { int cas = 0; scanf("%d", &t); while (t--) { scanf("%d", &n); printf("Case %d: %I64d\n", ++cas, solve()); } return 0;}
F:
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int N = 205;int x, y, n;bool vis[N][N][2];struct State { int x, y, flag, d; State() {} State(int x, int y, int flag, int d) { this->x = x; this->y = y; this->flag = flag; this->d = d; }};int solve() { if (y > x && x) return -1; memset(vis, false, sizeof(vis)); queue<State> Q; vis[x][y][0] = true; Q.push(State(x, y, 0, 0)); while (!Q.empty()) { State u = Q.front(); Q.pop(); if (u.x == 0 && u.y == 0) return u.d; int yang[2], lang[2]; yang[0] = u.x; yang[1] = x - u.x; lang[0] = u.y; lang[1] = y - u.y; for (int i = 0; i <= n && i <= lang[u.flag]; i++) { for (int j = 0; j + i <= n && j <= yang[u.flag]; j++) { if (i + j == 0) continue; if (lang[u.flag] - i > yang[u.flag] - j && yang[u.flag] - j) continue; if (lang[!u.flag] + i > yang[!u.flag] + j && yang[!u.flag] + j) continue; if (i > j && j) continue; State v; if (u.flag) v.x = yang[0] + j, v.y = lang[0] + i; else v.x = yang[0] - j, v.y = lang[0] - i; v.flag = !u.flag; v.d = u.d + 1; if (vis[v.x][v.y][v.flag]) continue; vis[v.x][v.y][v.flag] = true; Q.push(State(v.x, v.y, v.flag, v.d)); } } } return -1;}int main() { while (~scanf("%d%d%d", &x, &y, &n)) { printf("%d\n", solve()); } return 0;}
G:
#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int N = 205;int x, y, m, n;bool vis[N][N][2];struct State { int x, y, flag, d; State() {} State(int x, int y, int flag, int d) { this->x = x; this->y = y; this->flag = flag; this->d = d; }};int solve() { if (x > 20) return -1; if (y > x && x) return -1; memset(vis, false, sizeof(vis)); queue<State> Q; vis[x][y][0] = true; Q.push(State(x, y, 0, 0)); while (!Q.empty()) { State u = Q.front(); Q.pop(); if (u.x == 0 && u.y == 0) return u.d; int yang[2], lang[2]; yang[0] = u.x; yang[1] = x - u.x; lang[0] = u.y; lang[1] = y - u.y; for (int i = 0; i <= n && i <= lang[u.flag]; i++) { for (int j = 0; j + i <= n && j <= yang[u.flag]; j++) { if (i + j == 0) continue; if (lang[u.flag] - i > yang[u.flag] - j && yang[u.flag] - j) continue; if (lang[!u.flag] + i > yang[!u.flag] + j && yang[!u.flag] + j) continue; if (i > j && j) continue; State v; if (u.flag) v.x = yang[0] + j, v.y = lang[0] + i; else v.x = yang[0] - j, v.y = lang[0] - i; v.flag = !u.flag; v.d = u.d + 1; if (vis[v.x][v.y][v.flag]) continue; vis[v.x][v.y][v.flag] = true; Q.push(State(v.x, v.y, v.flag, v.d)); } } } return -1;}int main() { while (~scanf("%d%d", &m, &n)) { if (n <= 3) { x = y = m; printf("%d\n", solve()); } else { int ans = 0; while (m - ((n / 2) - 1) > n) { m -= (n / 2 - 1); ans += 2; } if (m > n) { if (n % 2) { if (n >= m - n / 2) { ans += 5; printf("%d\n", ans); continue; } } } while (m > n) { m -= (n / 2 - 1); ans += 2; } if (m * 2 <= n) ans += 1; else if (m == n) ans += 5; else ans += 3; printf("%d\n", ans); } } return 0;}
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