Codeforces Round #394 (Div. 2)
传送门：http://codeforces.com/contests/761
A题：[l,r]中有a个偶数，b个奇数，问存不存在l和r，1<=l<=r。很容易想到，如果abs(a-b)<=1，那么l和r就是存在的，除了一种情况，就是a=b=0的时候。由于l和r都大于等于1，所以当a=b=0时，不存在这种情况
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <string> #include <stack> #include <map> #include <set> #include <bitset> #define X first #define Y second #define clr(u,v); memset(u,v,sizeof(u)); #define in() freopen("data","r",stdin); #define out() freopen("ans","w",stdout); #define Clear(Q); while (!Q.empty()) Q.pop(); #define pb push_back using namespace std; typedef long long ll; typedef pair<int, int> pii; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; int main() { int a, b; cin >> a >> b; if (a == 0 || b == 0) puts("NO"); else if (b == a || b == a + 1 || a == b + 1) puts("YES"); else puts("NO"); return 0; }
B题：有n个障碍，还有L长度的圈。分别给出A和B遇到这n个障碍的时间，问他们是否在同一个轨道上。直接处理出时间差，然后对时间差不断旋转，看是否匹配。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <string> #include <stack> #include <map> #include <set> #include <bitset> #define X first #define Y second #define clr(u,v); memset(u,v,sizeof(u)); #define in() freopen("data","r",stdin); #define out() freopen("ans","w",stdout); #define Clear(Q); while (!Q.empty()) Q.pop(); #define pb push_back using namespace std; typedef long long ll; typedef pair<int, int> pii; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; int N[2][210]; vector <int> temp, temp2; int main() { int n, L; cin >> n >> L; for (int i = 0; i < n; i++) cin >> N[0][i]; for (int i = 0; i < n; i++) cin >> N[1][i]; int num = L; for (int i = 1; i < n; i++) { temp.pb(N[0][i] - N[0][i-1]); num -= N[0][i] - N[0][i-1]; } temp.pb(num); num = L; for (int i = 1; i < n; i++) { temp2.pb(N[1][i] - N[1][i-1]); num -= N[1][i] - N[1][i-1]; } temp2.pb(num); int sz = temp.size(); for (int j = 0; j < sz; j++) { int flag = 1; for (int i = 0; i < sz; i++) { if (temp2[i] != temp[(j+i)%sz]) { flag = 0; break; } } if (flag) { puts("YES"); return 0; } } puts("NO"); return 0; }
C题：给你n个字符串，每个字符串长度都为m，然后有n个光标，每个光标在每个字符串的开头，光标可以左右移动，问光标最少移动多少次，才能光标位置上的字符组合成的字符串包含至少一个符号，一个字母，一个数字。只要把每个字符串的光标移动后得到符号、字母、数字的最小次数预处理出来，然后记录下id，再排序，然后3个for暴力找最优解。只需要找每项的前三个就好了。这种是贪心的方法，也可以不排序，直接暴力3个for查询全部，不过复杂度是n3，也可以过。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <string> #include <stack> #include <map> #include <set> #include <bitset> #define X first #define Y second #define clr(u,v); memset(u,v,sizeof(u)); #define in() freopen("data","r",stdin); #define out() freopen("ans","w",stdout); #define Clear(Q); while (!Q.empty()) Q.pop(); #define pb push_back using namespace std; typedef long long ll; typedef pair<int, int> pii; const int maxn = 1e5 + 10; const int INF = 10100; char str[110][110]; pii F[110], N[110], C[110]; int main() { int n, len; cin >> n >> len; for (int i = 0; i < n; i++) cin >> str[i]; for (int i = 0; i < n; i++) { F[i].X = N[i].X = C[i].X = INF; F[i].Y = N[i].Y = C[i].Y = i; for (int j = 0; j < len; j++) { int pos = min(j, len - j); if (str[i][j] >= '0' && str[i][j] <= '9') N[i].X = min(N[i].X, pos); else if (str[i][j] >= 'a' && str[i][j] <= 'z') C[i].X = min(C[i].X, pos); else F[i].X = min(F[i].X, pos); } } sort(F, F + n); sort(N, N + n); sort(C, C + n); int ans = INF; for (int i = 0; i < 3; i++) for (int j = 0; j < 3; j++) for (int k = 0; k < 3; k++) { if (N[i].Y != F[j].Y && F[j].Y != C[k].Y && N[i].Y != C[k].Y) ans = min(ans, N[i].X + F[j].X + C[k].X); } cout << ans << endl; return 0; }
D题：给你n个数，和一个范围l和r，接下来有数组a[n]，下一行是数组p[n]，假设有数组c[n]，他按数字大小的排名对应p[n]，也可以看做p[n]是c[n]的离散化。问存不存在数组b，使得c[i]=b[i]-a[i],且b[i]的范围在[l,r]中。可以先对p排序，然后从小到大处理出c的值，我们假设p={1，2，3，4，5}，那么c[0]一定等于l，c[i]=max(l,c[i-1]-a[i-1]+a[i]+1);下标要按离散化的下标进行操作。
#include <iostream> #include <stdio.h> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <string> #include <stack> #include <map> #include <set> #include <bitset> #define X first #define Y second #define clr(u,v); memset(u,v,sizeof(u)); #define in() freopen("data","r",stdin); #define out() freopen("ans","w",stdout); #define Clear(Q); while (!Q.empty()) Q.pop(); #define pb push_back using namespace std; typedef long long ll; typedef pair<int, int> pii; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; int a[maxn]; pii p[maxn]; int ans[maxn]; int main() { // ios::sync_with_stdio(false); int n, l, r; scanf("%d%d%d", &n, &l, &r); for (int i = 0; i < n; i++) scanf("%d", &a[i]); for (int i = 0; i < n; i++) { scanf("%d", &p[i].X); p[i].Y = i; } sort(p, p + n); int pos = l; ans[p[0].Y] = l; for (int i = 1; i < n; i++) { pos = max(l, pos + a[p[i].Y] - a[p[i-1].Y] + 1); ans[p[i].Y] = pos; if (pos > r) { puts("-1"); return 0; } } for (int i = 0; i < n; i++) printf("%d ", ans[i]); puts(""); return 0; }
E题：给你一棵节点为n的树，让你在平面直角坐标系上画出这样一棵树，他的边必须平行或垂直于x轴，边长不限，点坐标绝对值不超过1e18。问能否构造出这棵树，且边不相交，如果能，输出点的坐标。这题要构造出一组边长序列，用N[i]代表边长，sum[i]代表N[i]~N[n-1]的和（n个点的树有n-1条边），那么N[i]必须要大于sum[i+1]。可以考虑，当N为2的幂的时候，满足N的要求。（2>1,4>1+2,8>1+2+4），于是边长序列就构造出来了。我们从边长为1<<n开始，一路dfs下去，上下左右递归，每个点最多有3个子节点和一个父节点（超过输出NO），递归后边长/2，这样就保证了边长不会相交，至于为什么，可以自己画一下图。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <string> #include <stack> #include <map> #include <set> #include <bitset> #define X first #define Y second #define clr(u,v); memset(u,v,sizeof(u)); #define in() freopen("data","r",stdin); #define out() freopen("ans","w",stdout); #define Clear(Q); while (!Q.empty()) Q.pop(); #define pb push_back using namespace std; typedef long long ll; typedef pair<int, int> pii; const int maxn = 1e3 + 10; const int INF = 0x3f3f3f3f; int head[maxn], cnt , vis[maxn], flag; pii ans[maxn]; const int Next[4][2] = {{1, 0}, {0, 1}, {0, -1}, { -1, 0}}; struct Edge { int v, to; } E[maxn]; void init() { clr(head, -1); cnt = 0; } void addedge(int u, int v) { E[cnt].v = u, E[cnt].to = head[v]; head[v] = cnt++; } void dfs(int cur, int x, int y, int L, int k) { if (vis[cur]) return ; vis[cur] = 1; ans[cur] = make_pair(x, y); int pos = 0, sum = 0; for (int i = head[cur]; ~i; i = E[i].to) { sum++; int v = E[i].v; if (vis[v]) continue; if (pos + k == 3) pos++; dfs(v, x + Next[pos][0]*L, y + Next[pos][1]*L, L >> 1, pos); pos++; } if (sum > 4) flag = 1; } int main() { init(); int n; cin >> n; for (int i = 1; i < n; i++) { int u, v; cin >> u >> v; addedge(u, v), addedge(v, u); } dfs(1, 0, 0, 1 << 30, 4); if (flag) cout << "NO" << endl; else { cout << "YES" << endl; for (int i = 1; i <= n; i++) cout << ans[i].X << " " << ans[i].Y << endl; } return 0; }
2017-02-02 21:11:39
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