2014 北京邀请赛ABDHJ题解
A. A Matrix
构造，结论是从第一行開始往下产生一条曲线，使得这条区间最长且从上到下递减，
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stdio.h>
#include <vector>
#include <set>
using namespace std;
#define N 100005
vector<int>G[N], P[N], tmp;
set<int>s[N];
set<int>::iterator it;
int n,m;
bool work(){
for(int i = m; i>1; i--) {
for(int j = G[i].size()-1; j>=0; j--) {
it = s[i-1].lower_bound(-G[i][j]);
if(it==s[i-1].end())return false;
int pos = lower_bound(G[i-1].begin(), G[i-1].end(), -(*it))-G[i-1].begin();
P[i-1][pos] = j;
s[i-1].erase(it);
}
}
return true;
}
void init(){
for(int i = 0; i <= n; i++)G[i].clear(), s[i].clear(), P[i].clear();
}
int Stack[N];
int main(){
int T, Cas = 1,i,j,k;cin>>T;
while(T--) {
init();
scanf("%d %d",&n,&m);
bool success = true;
for(i=1;i<=m;i++){
scanf("%d",&k);
for(int z = 0; z < k; z++)
{
scanf("%d",&j);
G[i].push_back(j);
s[i].insert(-j);
if(z && G[i][z-1]>G[i][z])
success=false;
P[i].push_back(-1);
}
}
printf("Case #%d: ",Cas++);
if(success && work()) {
int top = 0;
for(i = 0; i < P[1].size(); i++)
{
int h = 1, l = i;
tmp.clear();
while(1)
{
tmp.push_back(G[h][l]);
if(P[h][l]==-1)break;
l = P[h][l];
h++;
}
for(j=tmp.size()-1;j>=0;j--)
Stack[top++] = tmp[j];
}
for(i=0;i<top;i++)
printf("%d%c",Stack[i],i==top-1?'\n':' ');
}
else puts("No solution");
}
return 0;
}
B. Beautiful Garden
暴力
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
const int N = 42;
int n, a[N];
int dis[N * N], dep;
bool vis[N][N];
int main() {
int T;
scanf("%d", &T);
for (int cas = 1; cas <= T; ++cas) {
scanf("%d", &n);
for (int i = 0; i < n; ++i)
scanf("%d", &a[i]);
std::sort(a, a + n);
dep = 0;
for (int i = 0; i < n; ++i)
for (int j = i + 1; j < n; ++j)
dis[dep++] = a[j] - a[i];
std::sort(dis, dis + dep);
dep = std::unique(dis, dis + dep) - dis; int ans = n - 1, cnt, idx, f;
ll st; for (int i = 0; i < dep; ++i) {
memset(vis, 0, sizeof vis);
for (int k = 0; k < n; ++k)
for (int z = 0; z < n; ++z)
if(!vis[k][z]) {
st = a[k] - (ll)z * dis[i];
cnt = idx = 0;
f = 1;
for (int l = 0; l < n; ++l) {
while (idx < n && a[idx] < st)
++idx;
if (idx < n && st == a[idx]) {
if (vis[idx][l]) {
f = 0;
break;
}
vis[idx][l] = true;
++cnt, ++idx;
}
st += dis[i];
}
if (f && n - cnt < ans)
ans = n - cnt;
}
}
printf("Case #%d: %d\n", cas, ans);
} return 0;
}
E. Elegant String
先得到一个dp方程
dp[i][j]表示字符串长度为i，结尾有j个数互不同样的方法数
然后得到转移方程再用矩阵高速幂加速
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <stdio.h>
#include <vector>
#include <set>
using namespace std;
#define ll long long
#define Matr 11
#define mod 20140518
struct mat{
ll a[Matr][Matr],size;
mat(){
size = 0;
memset(a, 0, sizeof a);
}
};
mat multi(mat m1,mat m2){
mat ans = mat(); ans.size = m1.size;
for(int i = 1; i <= m1.size; i++)
for(int j = 1; j<= m2.size; j++)
if(m1.a[i][j])
for(int k = 1; k <= m1.size; k++)
ans.a[i][k] = (ans.a[i][k]+m1.a[i][j]*m2.a[j][k])%mod;
return ans;
}
mat quickmulti(mat m,ll n){
mat ans = mat();
ll i;
for(i=1;i<=m.size;i++)ans.a[i][i] = 1;
ans.size = m.size;
while(n){
if(n&1)ans = multi(m,ans);
m=multi(m,m);
n /= 2;
}
return ans;
}
ll n,k;
int main(){
ll T, Cas = 1;cin>>T;
while(T--){
cin>>n>>k;
mat z = mat();
for(ll i = 1; i <= k; i++)
for(ll j = i; j <= k; j++)
z.a[i][j] = 1;
ll now = k;
for(ll i = 2; i <= k; i++, now--)
z.a[i][i-1]+=now;
z.size = k;
z = quickmulti(z, n-1);
ll tmp = 0;
for(ll i = 1; i <= k; i++)
tmp += z.a[i][1];
tmp *= (k+1);
tmp %= mod;
printf("Case #%lld: %lld\n",Cas++,tmp);
}
return 0;
}
H. Happy Reversal
把全部数字取法后都一起排序，最大的减最小的就可以
可能最大最小都来源一个数，则答案就是 第二大减最小 或者最大减第二小
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 20005;
ll a[N], b[N];
ll er[N];
int main() {
int T;
char s[66];
er[0] = 1;
for(int i = 1;i<63;i++)er[i] = er[i-1]*2;
while(~scanf("%d", &T)) {
for(int cas = 1; cas <= T; cas ++) {
int n, m;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i ++) {
scanf("%s", s);
ll suma = 0;
ll sumb = 0;
for(int j = 0; j < m; j ++) {
suma = suma * 2 + (s[j] == '1');
sumb = sumb * 2 + (s[j] == '0');
}
a[i*2] = max(suma, sumb);
a[i*2+1] = min(suma, sumb);
}
sort(a, a + n + n);
ll ans;
if( a[2*n-1] + a[0] + 1 == er[m] ) ans = max(a[2*n-2] - a[0], a[2*n-1] - a[1]);
else ans = a[2*n-1] - a[0];
printf("Case #%d: %lld\n", cas, ans);
}
}
return 0;
}
J. Justice String
J：点击打开链接
hash二分第一个不同的字符距离0的位置，再二分第二个不同的字符距离第一个字符的位置
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX_N = 3000007;
int sa[MAX_N], ws[MAX_N], rank[MAX_N], height[MAX_N];
#define F(x) ((x)/3+((x)%3==1? 0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
int wa[MAX_N],wb[MAX_N],wv[MAX_N],wss[MAX_N];
struct suffix {
int c0(int *r, int a, int b) {
return r[a] == r[b] && r[a+1] == r[b+1] && r[a+2] == r[b+2];
}
int c12(int k,int *r,int a,int b) {
if(k == 2)
return r[a] < r[b] || ( r[a] == r[b] && c12(1,r,a+1,b+1) );
else return r[a] < r[b] || ( r[a] == r[b] && wv[a+1] < wv[b+1] );
}
void sort(int *r,int *a,int *b,int n,int m) {
int i;
for(i = 0;i < n;i++)wv[i] = r[a[i]];
for(i = 0;i < m;i++)wss[i] = 0;
for(i = 0;i < n;i++)wss[wv[i]]++;
for(i = 1;i < m;i++)wss[i] += wss[i-1];
for(i = n-1;i >= 0;i--)
b[--wss[wv[i]]] = a[i];
}
void dc3(int *r,int *sa,int n,int m) {
int i, j, *rn = r + n;
int *san = sa + n, ta = 0, tb = (n+1)/3, tbc = 0, p;
r[n] = r[n+1] = 0;
for(i = 0;i < n;i++)if(i %3 != 0)wa[tbc++] = i;
sort(r + 2, wa, wb, tbc, m);
sort(r + 1, wb, wa, tbc, m);
sort(r, wa, wb, tbc, m);
for(p = 1, rn[F(wb[0])] = 0, i = 1;i < tbc;i++)
rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p - 1 : p++;
if(p < tbc)dc3(rn,san,tbc,p);
else for(i = 0;i < tbc;i++)san[rn[i]] = i;
for(i = 0;i < tbc;i++) if(san[i] < tb)wb[ta++] = san[i] * 3;
if(n % 3 == 1)wb[ta++] = n - 1;
sort(r, wb, wa, ta, m);
for(i = 0;i < tbc;i++)wv[wb[i] = G(san[i])] = i;
for(i = 0, j = 0, p = 0;i < ta && j < tbc;p++)
sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
for(;i < ta;p++)sa[p] = wa[i++];
for(;j < tbc;p++)sa[p] = wb[j++];
}
void DA(int str[],int sa[],int n,int m) {
for(int i = n;i < n*3;i++)
str[i] = 0;
dc3(str, sa, n+1, m);
int i,j,k = 0;
for(i = 0;i <= n;i++)rank[sa[i]] = i;
for(i = 0;i < n; i++)
{
if(k) k--;
j = sa[rank[i]-1];
while(str[i+k] == str[j+k]) k++;
height[rank[i]] = k;
}
}
};
int RMQ[MAX_N], mm[MAX_N], best[20][MAX_N];
void initRMQ(int n) {
mm[0] = -1;
for(int i = 1; i <= n; ++i) {
mm[i]=((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
}
for (int i = 1; i <= n; ++i) {
best[0][i] = i;
}
for (int i = 1; i <= mm[n]; ++i) {
for (int j = 1; j + (1 << i) - 1 <= n; ++j) {
int a = best[i - 1][j];
int b = best[i - 1][j + (1 << (i - 1))];
if (RMQ[a] < RMQ[b]) best[i][j] = a;
else best[i][j] = b;
}
}
}
int query(int a, int b) {
int t = mm[b - a + 1];
b -= (1 << t) - 1;
a = best[t][a], b = best[t][b];
return RMQ[a] < RMQ[b] ? a : b;
}
int lcp(int a, int b) {
a = rank[a], b = rank[b];
if (a > b) swap(a, b);
return height[query(a + 1, b)];
}
char A[MAX_N],B[MAX_N];
int r[MAX_N], da[MAX_N];
int main() {
int T;
int cas = 0;
scanf("%d",&T);
while(T-- > 0) {
suffix ar;
scanf("%s%s",A, B);
int len1 = strlen(A);
int len2 = strlen(B);
if(len1 < len2) {
printf("Case #%d: -1\n", ++cas);
continue;
}
int n = len1 + len2 + 1;
for(int i = 0; i < len1; ++i) r[i] = A[i];
for(int i = 0; i < len2; ++i) r[len1 + 1 + i] = B[i];
r[len1] = 1;
r[n] = 0;
ar.DA(r, sa, n, 128); for(int i = 1; i <= n; ++i)
RMQ[i] = height[i];
initRMQ(n);
int ans = -1;
for(int i = 0; i <= len1 - len2; ++i) {
int st = i;
int ed = len1 + 1;
int tmp = lcp(st, ed);
// printf("here1 %d %d %d\n", st, ed, tmp);
st += tmp;
ed += tmp;
if(ed >= n) {
ans = i;
break;
}
st++;
ed++;
if(ed >= n) {
ans = i;
break;
}
tmp = lcp(st, ed);
// printf("here2 %d %d %d\n", st, ed, tmp);
st += tmp;
ed += tmp;
if(ed >= n) {
ans = i;
break;
}
st++;
ed++;
if(ed >= n) {
ans = i;
break;
}
tmp = lcp(st, ed);
// printf("here3 %d %d %d\n", st, ed, tmp);
st += tmp;
ed += tmp;
if(ed >= n) {
ans = i;
break;
}
}
printf("Case #%d: %d\n", ++cas, ans);
}
return 0;
}
2014 北京邀请赛ABDHJ题解的更多相关文章
- 2014 ACM/ICPC 北京邀请赛 部分 题解
题目链接:http://acm.bnu.edu.cn/bnuoj/problem.php?search=2014+ACM-ICPC+Beijing+Invitational+Programming+C ...
- hihocoder 1084 扩展KMP && 2014 北京邀请赛 Justice String
hihocoder 1084 : http://hihocoder.com/problemset/problem/1084 北京邀请赛 Just String http://www.bnuoj.co ...
- BNUOJ 34985 Elegant String 2014北京邀请赛E题 矩阵快速幂
题目链接:http://acm.bnu.edu.cn/bnuoj/problem_show.php?pid=34985 题目大意:问n长度的串用0~k的数字去填,有多少个串保证任意子串中不包含0~k的 ...
- 2014 北京 DevFest 大会能够报名啦，小伙伴们还在等什么
一年一度的大型开发人员活动,2014 北京 DevFest 大会站点正式上线: http://devfest.gdgbeijing.org/. 还等什么,開始报名了! 今年 DevFest 大会将再次 ...
- bnu A Matrix 北京邀请赛A题
A Matrix Time Limit: 2000ms Memory Limit: 65536KB 64-bit integer IO format: %lld Java class n ...
- HDU5092——Seam Carving(动态规划+回溯)(2014上海邀请赛重现)
Seam Carving DescriptionFish likes to take photo with his friends. Several days ago, he found that s ...
- HDU5093——Battle ships(最大二分匹配)(2014上海邀请赛重现)
Battle ships Problem DescriptionDear contestant, now you are an excellent navy commander, who is res ...
- HDU5099——Comparison of Android versions(简单题)(2014上海邀请赛重现)
Comparison of Android versionsProblem DescriptionAs an Android developer, itˇs really not easy to fi ...
- HDU5090——Game with Pearls(匈牙利算法|贪心)(2014上海邀请赛重现)
Game with Pearls Problem DescriptionTom and Jerry are playing a game with tubes and pearls. The rule ...
随机推荐
- BSTR、char*和CString转换
(1) char*转换成CString 若将char*转换成CString,除了直接赋值外,还可使用CString::Format进行.例如: char chArray[] = "This ...
- 《码农周刊》干货精选（Python 篇）
<码农周刊>已经累计发送了 38 期,我们将干货内容进行了精选.此为 Python 篇. <码农周刊>往期回顾:http://weekly.manong.io/issues/ ...
- La=LaULb （单链表）
#include<stdio.h> typedef struct LNode { int data; struct LNode *next; }LNode,*LinkList; void ...
- 【转】VPN服务器配置详解
参考博文: VPN服务器配置详解 等公司上服务器开始配置 vpn
- python自学笔记（九）python练习题
1. 已知字符串 a = "aAsmr3idd4bgs7Dlsf9eAF",要求如下 1.1 请将a字符串的大写改为小写,小写改为大写 print a.swapcase() 1.2 ...
- asp.net mvc重写RequestValidator
/// <summary> /// <httpRuntime requestValidationType="xxx.CustomRequestValidator" ...
- Lotus Sametime 服务器的安装和配置
IBM Lotus Sametime 是一款强大的实时协作软件,目前最新版本是 7.5.1.通过它,您不仅能够进行网络聊天,而且可以方便地召开网络会议.在网络社区中与其他人进行沟通.了解更多关于 Lo ...
- 英文论文中“such as, for example, e.g., i.e., etc., et al. ”的用法分析 (转)
在英文论文的编辑加工中,常会遇到such as, for example, e. g. , i. e. , etc. 和et al. 的错误及混淆使用.这里,举例分析这几个词的意义,并阐述其正确用法. ...
- android studio 快捷笔记
setting->editor->打勾 ctrl+Q ctrl+tab alt+回车 ctrl+shift+回车
- javascript 字符串方法传参
javascript 字符串方法传参由于嵌套的单引号,双引号过多.有点混乱.. 正确方法如下: ' <td align="left"><input type= ...