2014 ACM/ICPC Asia Regional Xi'an Online
2014西安网络赛
A. Post Robot
把每种单词都kmp跑一遍，顺序输出即可
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
inline int read() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
const int N = 1000500;
using namespace std;
int n;
string s,t,s1="Apple",s2="iPhone",s3="iPod",s4="iPad",s5="Sony";
int nxt[N],vis[N];
void kmp_pre(string s) {
int i,j,m=s.size();
j=nxt[0]=-1;
i=0;
while(i<m){
while(-1!=j&&s[i]!=s[j])j=nxt[j];
nxt[++i]=++j;
}
}
void kmp(string s,string t,int f) {
int i,j,n=s.size(),m=t.size();
i=j=0;
while(i<n){
while(-1!=j&&s[i]!=t[j])j=nxt[j];
++i;++j;
if(j>=m) {
//printf("%d ",i);
vis[i] = f;
j=nxt[j];
}
}
}
int main() {
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
memset(vis,-1,sizeof(vis));
int f=0;
while(getline(cin,t)){
if(f) s+=" ";f=1;
s+=t;
}
kmp_pre(s1); kmp(s,s1,0);
kmp_pre(s2); kmp(s,s2,0);
kmp_pre(s3); kmp(s,s3,0);
kmp_pre(s4); kmp(s,s4,0);
kmp_pre(s5); kmp(s,s5,1);
rep(i,1,s.size())
if(vis[i]==0) puts("MAI MAI MAI!");
else if(vis[i]==1) puts("SONY DAFA IS GOOD!");
return 0;
}
B. Boring String Problem
后缀数组的帮我们排好序了，然后取后缀的前缀即可。严格第k小，就处理一下减去Height[i]即可去重，取前缀和，二分就可以找到严格第k小的串，然后问题就是，求一个串它在整个串中出现的最左边的位置。以当前这个串的左端点为中心，直接在Height数组里二分最左或最右的公共前缀长度恰好大于等于串长的位置，然后在这个区间询问最小的SA值即可。讲道理正解挺好想的。。。（用了两天发现自己的后缀数组板子有错，然后又发现不会二分。。。网上又找了一份大佬板子
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
const int N = 100005;
inline int readint() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
using namespace std;
int n;
char c[N],str[N];
int rnk[N] , SA[N] , Height[N];
int X[N] , Y[N] , sum[N];
int f[101000][20] , fm[101000][20];
bool cmp(int *r,int a,int b,int l) {
return ( r[a] == r[b] && r[a+l] == r[b+l] );
}
void calc() {
int l , p , *x = X , *y = Y , m = 128;
rep(i,0,m) sum[i] = 0;
rep(i,1,n) sum[ x[i] = c[i] ] ++;
rep(i,1,m) sum[i] += sum[i-1];
per(i,n,1) SA[ sum[ x[i] ]-- ] = i;
for ( l = 1 , p = 1 ; l <= n ; m = p , l *= 2 ) {
p = 0;
rep(i,n-l+1,n) y[++p] = i;
rep(i,1,n) if ( SA[i] > l ) y[++p] = SA[i] - l;
rep(i,0,m) sum[i] = 0;
rep(i,1,n) sum[ x[y[i]] ] ++;
rep(i,1,m) sum[i] += sum[i-1];
per(i,n,1) SA[ sum[ x[y[i]] ]-- ] = y[i];
swap( x , y );
x[SA[1]] = 1; p = 2;
rep(i,2,n)
x[ SA[i] ] = cmp(y,SA[i-1],SA[i],l) ? p - 1 : p++;
}
rep(i,1,n) rnk[SA[i]] = i;
p = 0;
rep(i,1,n) {
if ( rnk[i] == 1 ) continue;
while ( c[i+p] == c[SA[rnk[i]-1]+p] ) p ++;
Height[rnk[i]] = p;
if ( p ) p --;
}
}
void init() {
n = strlen(str);
for(int i=0;i<n;++i) c[i+1] = str[i];c[n+1]=0;
}
int ST_h[N][22], ST_sa[N][22],Log[N];
void init_rmq() {
Log[1] = 0;
rep(i,2,n) Log[i]=Log[i>>1]+1LL;
rep(i,1,n) ST_h[i][0]=Height[i], ST_sa[i][0]=SA[i];
rep(j,1,20)for(int i=1;i+(1<<j-1)<=n;++i) {
ST_h[i][j] = min(ST_h[i][j-1],ST_h[i+(1<<(j-1))][j-1]);
ST_sa[i][j] = min(ST_sa[i][j-1],ST_sa[i+(1<<(j-1))][j-1]);
}
}
int ask_h(int l,int r) {
if(l==r)return ST_h[l][0];
if(l>r)swap(l,r);
++l;
int p = Log[r-l+1];
return min(ST_h[l][p],ST_h[r-(1<<p)+1][p]);
}
int ask_sa(int l,int r) {
if(l>r)swap(l,r);
int p = Log[r-l+1];
return min(ST_sa[l][p],ST_sa[r-(1<<p)+1][p]);
}
ll s[N];
int fd(int L,int len) {
int l1=1,r1=L-1,ansl=L;
while(l1<r1) {
int mid = (l1+r1)>>1;
if(ask_h(mid,L)>=len) r1=mid;
else l1=mid+1;
}
if(ask_h(l1,L)>=len)ansl = min(ansl,l1);
int l2=L+1,r2=n,ansr=L,tmp=-1;
while(l2<=r2) {
int mid = (l2+r2)>>1;
if(ask_h(L,mid)<len) tmp=mid,r2=mid-1;
else l2=mid+1;
}
if(tmp!=-1&&ask_h(L,tmp)<len)--tmp;
else if(tmp==-1&&ask_h(L,n)>=len) tmp=n;
if(tmp>=L+1&&tmp<=n&&ask_h(L,r2)>=len)ansr = max(r2,ansr);
int x = ask_sa(ansl,ansr);
return x;
}
void solve(ll k,int &l,int &r) {
int p = lower_bound(s+1,s+1+n,k) - s;
l = SA[p];
int len = k - s[p-1] + Height[p];
l = fd(p,len);
r = l+len-1;
return;
}
int main() {
while(scanf(" %s",str)!=EOF) {
init();
calc();
init_rmq();
sum[0]=0;
rep(i,1,n) s[i] = s[i-1] + (n-SA[i]-Height[i]+1);
int q=readint();
int l=0,r=0;
while(q--) {ll x;
scanf("%I64d",&x);
x = (ll)(l^r^x) + 1LL;
if(x>(ll)s[n]) {
l=r=0;
puts("0 0");
continue;
}
solve(x,l,r);
printf("%d %d\n",l,r);
}
}
return 0;
}
C. Paint Pearls
先想暴力dp，\(dp[i] = min(dp[j-1]+D(j,i)^2), D(l,r)：表示[l,r]区间内不同数的个数\)。长得想斜率优化，可是搞不出来。然后，发现只有区间的长度大于区间种类的平方时，才会有贡献，否则不如一个个加起来。于是枚举以i为右端点的区间包含k种数字，显然\(k^2\)不能大过i，现在就是要求以i为右端点包含k种数字，的最左边的位置，一开始写了主席树，然后二分左端点的两个log做法。tle了之后，发现可以双指针预处理pre[i][j]表示以i为右端点，包含j种元素的最远左端点，因为有个离散化，是一个log。所以预处理复杂度是\(O(n \sqrt{n} logn)\)，dp是\(O(n \sqrt{n})\)的，然而还是t了。。。可能这个复杂度还是有点问题。。。还是写搓了？？我再努力补。。
不优化确实会tle，学习了一些优化，换成刷表法dp，最重要优化的是当计算出的dp[i]大于等于dp[n]时break。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define mem(W) memset(W,0,sizeof(W))
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
const int N = 50010;
const ll inf = 1000000000000000LL;
inline int read() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
using namespace std;
vector<ll> v;
int getid(ll x) {
return lower_bound(v.begin(),v.end(),x) - v.begin() + 1;
}
int n;
int a[N],b[N],A[N];
ll dp[N];
int vis[N];
int main() {
while(scanf("%d",&n)!=EOF) {
v.clear();
rep(i,1,n)a[i]=read(),dp[i]=inf,v.pb(a[i]);
sort(v.begin(),v.end());
v.erase(unique(v.begin(),v.end()),v.end());
for(int i=1;i<=n;++i) b[i]=getid(a[i]);
dp[0]=0;
rep(i,0,n-1) {
int sz = 0;
rep(j,i+1,n) {
if(!vis[b[j]]) vis[b[j]]=1, A[++sz]=b[j];
if(dp[i]+sz*sz>=dp[n]) break; //***
if(sz*sz>n) break;
dp[j] = min(dp[j],dp[i]+sz*sz);
}
rep(j,1,sz) vis[A[j]]=0;
}
printf("%I64d\n",dp[n]);
}
return 0;
}
E. Game
手推了几组，找规律的。把每一堆的大小都异或起来就星了。
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
inline int read() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
using namespace std;
int n,x;
int main() {
while(scanf("%d",&n)!=EOF){
int ans=0;
rep(i,1,n){
int x=read();ans^=x;
}
if(!ans)puts("Lose");
else puts("Win");
}
return 0;
}
F. Dice
直接从123456开始bfs出到每种状态的距离即可。查询时，替换下标即可。
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
const int N = 654321 + 10;
inline int read() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
using namespace std;
int hs(string s) {
int hs = 0;
for(int i=0;i<6;++i) hs = hs*10 + (s[i]-'0');
return hs;
}
string t;
string fhs(int x) {
t.clear();
while(x) {
t += (char)(x%10+'0');
x/=10;
}
reverse(t.begin(),t.end());
return t;
}
string tx,tx2;
int rt1(int x) {
tx = fhs(x);
tx2 = tx;
tx2[2-1] = tx[3-1];
tx2[4-1] = tx[2-1];
tx2[1-1] = tx[4-1];
tx2[3-1] = tx[1-1];
return hs(tx2);
}
int rt2(int x) {
tx = fhs(x);
tx2 = tx;
tx2[2-1] = tx[4-1];
tx2[3-1] = tx[2-1];
tx2[1-1] = tx[3-1];
tx2[4-1] = tx[1-1];
return hs(tx2);
}
int rt3(int x) {
tx = fhs(x);
tx2 = tx;
tx2[2-1] = tx[5-1];
tx2[6-1] = tx[2-1];
tx2[1-1] = tx[6-1];
tx2[5-1] = tx[1-1];
return hs(tx2);
}
int rt4(int x) {
tx = fhs(x);
tx2 = tx;
tx2[2-1] = tx[6-1];
tx2[5-1] = tx[2-1];
tx2[1-1] = tx[5-1];
tx2[6-1] = tx[1-1];
return hs(tx2);
}
int dis[N],vis[N];
void bfs(string s) {
memset(dis,-1,sizeof(dis));
memset(vis,0,sizeof(vis));
int u = hs(s);
dis[u] = 0;
queue<int> q;
q.push(u);
vis[u]=1;
while(!q.empty()) {
int u = q.front(); q.pop();
int tu = u;
tu = rt1(u);
if(!vis[tu]) {
vis[tu] = 1;
q.push(tu);
dis[tu] = dis[u] + 1;
}
tu = rt2(u);
if(!vis[tu]) {
vis[tu] = 1;
q.push(tu);
dis[tu] = dis[u] + 1;
}
tu = rt3(u);
if(!vis[tu]) {
vis[tu] = 1;
q.push(tu);
dis[tu] = dis[u] + 1;
}
tu = rt4(u);
if(!vis[tu]) {
vis[tu] = 1;
q.push(tu);
dis[tu] = dis[u] + 1;
}
}
}
int n,a[10],b[10];
map<int,int> c;
int main() {
string s = "123456";
bfs(s);
while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6])!=EOF) {
scanf("%d%d%d%d%d%d",&b[1],&b[2],&b[3],&b[4],&b[5],&b[6]);
c.clear();
rep(i,1,6) c[a[i]] = i;
int hs = 0;
rep(i,1,6)hs=hs*10+c[b[i]];
printf("%d\n",dis[hs]);
}
return 0;
}
H. Number Sequence
一开始想起一道题，觉得要从高位到低位分治贪心。然后发现过的人有点多啊。就开始找规律。。。于是顺利浪费了大量时间。。。首先发现最大值就是\(n*(n+1)\) ，而每一个值都可以用一些\(2^i-1\)组合成，于是我们从大到小贪心的把每个数都异或成\(2^i-1\)中的尽可能大的值即可
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6 +100000;
inline int read() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
using namespace std;
ll n;
ll a[N],b[N];
int vis[N];
void solve() {
memset(vis,0,sizeof(vis));
for(int i=n;i>=0;--i) {
int e = 0;
for(ll j=20;j>=0;--j) {
e=(1<<j)-1;
if(!vis[(e^i)]&&(e^i)<=n) {
vis[(e^i)]=1;
b[i] = (e^i);
break;
}
}
}
}
int main() {
while(scanf("%lld",&n)!=EOF){
rep(i,0,n)a[i]=read();
solve();int f=0;
printf("%lld\n",(n+1ll)*n);
rep(i,0,n) {
if(f)printf(" ");f=1;
printf("%lld",b[a[i]]);
}puts("");
}
return 0;
}
I. 233 Matrix
发现n非常小，m很大，于是想到矩阵快速幂，推一下就ok了
#include <cstdio>
#include <algorithm>
#include <map>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
#include <iterator>
#include <string>
#include <deque>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define pb push_back
#define MP make_pair
#define fr first
#define sc second
#define PII pair<int,int>
#define VI vector<int>
#define rg register
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 10000007;
inline int read() {
char c=getchar();int x=0,f=1;
while(!isdigit(c)){if(f=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
using namespace std;
ll ans[20][20],c[20][20],d[20][20];
inline void mul(ll a[][20], ll b[][20], int n){
for(rg int i=1;i<=n;++i)
for(int j=1;j<=n;++j)ans[i][j]=0,c[i][j]=a[i][j],d[i][j]=b[i][j];
for(rg int i=1;i<=n;++i)
for(rg int k=1;k<=n;++k)if(c[i][k])
for(rg int j=1;j<=n;++j)if(d[k][j]){
ans[i][j] = (ans[i][j] + (c[i][k]*d[k][j]))%mod;
}
for(rg int i=1;i<=n;++i)
for(rg int j=1;j<=n;++j)a[i][j]=ans[i][j];
}
ll E[20][20],A[20][20];
inline void mx_pow(ll A[][20],ll b,int n) {
while(b) {
if(b&1) mul(E,A,n);
mul(A,A,n);
b>>=1LL;
}
}
int n;
ll m,a[20];
int main() {
while(scanf("%d%lld",&n,&m)!=EOF) {
memset(A,0,sizeof(A));
memset(E,0,sizeof(E));
rep(i,1,n) a[i] = read();
a[n+1] = 233; a[n+2] = 1;
rep(i,1,n+2)E[i][i] = 1;
rep(i,1,n)rep(j,1,i) A[i][j] = 1;
rep(i,1,n)A[i][n+1]=1;
A[n+1][n+1] = 10; A[n+1][n+2] = 3;
A[n+2][n+2]=1;
mx_pow(A,m,n+2);
ll res = 0;
rep(i,1,n+2) res = (res + (E[n][i]*a[i])%mod)%mod;
printf("%lld\n",res);
}
return 0;
}
