2014 西安网络赛部分题解
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1002 Boring String Problem
后缀数组+RMQ+二分
后缀数组二分确定第K不同子串的位置 , 二分LCP确定可选的区间范围 , RMQ求范围内最小的sa
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define prt(k) cout<<#k" = "<<k<<endl;
const int N = 100010;
int sa[N], rank[N], rank2[N], h[N],c[N], *x, *y, ans[N];
typedef long long ll;
char str[N];
bool cmp(int* r, int a, int b, int l, int n)
{
return r[a]==r[b] && a+l<n && b+l<n && r[a+l]==r[b+l];
}
void radix_sort(int n,int sz)
{
for(int i=0; i<sz; i++) c[i]=0;
for(int i=0; i<n; i++) c[x[y[i]]]++;
for(int i=1; i<sz; i++) c[i]+=c[i-1];
for(int i=n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];
}
void get_sa(char c[],int n,int sz=128)
{
x=rank,y=rank2;
for(int i=0;i<n;i++) x[i]=c[i],y[i]=i;
radix_sort(n,sz);
for(int len=1;len<n;len<<=1)
{
int yid=0;
for(int i=n-len;i<n;i++) y[yid++]=i;
for(int i=0;i<n;i++) if(sa[i]>=len) y[yid++]=sa[i]-len;
radix_sort(n,sz);
swap(x,y);
x[sa[0]]=yid=0;
for(int i=1;i<n;i++)
{
x[sa[i]]=cmp(y,sa[i],sa[i-1],len,n) ? yid : ++yid;
}
sz=yid+1;
if(sz>=n) break;
}
for(int i=0;i<n;i++) rank[i]=x[i];
}
void get_h(char str[],int n)
{
int k=0; h[0]=0x3f3f3f3f;
for(int i=0;i<n;i++)
{
if(rank[i]==0) continue;
k=max(k-1,0);
int j=sa[rank[i]-1];
while(i+k<n && j+k<n && str[i+k]==str[j+k]) k++;
h[rank[i]]=k;
}
}
int dp[N][20];
int mmm[N][22];
void RMQ_init(int n)
{
for(int i=0;i<n;i++) dp[i][0]=h[i], mmm[i][0]=sa[i];
dp[0][0]=0x3f3f3f3f;
for(int i=1;(1<<i)<=n;i++)
{
for(int j=0;j+(1<<i)-1<n; j++)
dp[j][i]=min(dp[j][i-1], dp[j+(1<<(i-1))][i-1]),
mmm[j][i]=min(mmm[j][i-1], mmm[j+(1<<(i-1))][i-1]);
}
}
int LCP(int l,int r,int n)
{
if(l==r) return n-sa[l];
l++;
if(l>r) swap(l,r);
int k = 0;
while(1<<(k+1) <= r-l+1) k++;
return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
ll Range[N];
int bin(ll x, int n)
{
int ans = -1;
int l=0, r=n-1, mid;
while(l<=r)
{
mid=(l+r)/2;
if(Range[mid]<x) ans=mid,l=mid+1;
else r=mid-1;
}
return ans;
}
int MMM(int l, int r)
{
if(l>r) swap(l,r);
int k = 0;
while(1<<(k+1) <= r-l+1) k++;
return min(mmm[l][k], mmm[r-(1<<k)+1][k]);
}
int binID(int x,int n,int len)
{
int ans=x;
int l=x,r=n-1,mid;
while(l<=r)
{
mid=(l+r)/2;
if(LCP(x,mid,n)>=len)
{
ans=mid;
l=mid+1;
}
else r=mid-1;
}
return ans;
}
int main()
{
while(scanf("%s",str)==1)
{
int n=strlen(str);
get_sa(str,n);
get_h(str,n);
RMQ_init(n);
h[0]=0;
for(int i=0;i<n;i++)
{
Range[i] = (n-sa[i])-h[i];
if(i-1>=0) Range[i]+=Range[i-1];
}
int q; cin>>q;
int L=0,R=0;
while(q--)
{
ll V; scanf("%I64d",&V);
ll K = (L^R^V) + 1;
if(K>Range[n-1])
{
L=0, R=0;
printf("0 0\n");
continue;
}
int id = bin(K,n);
ll jian=0;
if(id>=0) jian=Range[id];
ll res=K-jian;
id++;
int len=h[id]+res;
int hid=binID(id,n,len);
int Left=MMM(id,hid);
L=Left+1, R=Left+len;
printf("%d %d\n",L,R);
}
}
}
1003 Paint Pearls
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define prt(k) cout<<#k" = "<<k<<endl;
const int N = 50010;
#include<map>
#include<algorithm>
int a[N];
int n;
int dp[N];
int b[N];
int fa[N];
int f(int x) { return x==fa[x]?x:fa[x]=f(fa[x]); }
void U(int x, int y) { x=f(x),y=f(y); if(x-y) fa[y]=x; }
int sqr(int x) { return x * x; }
int last[N];
int main()
{
while(scanf("%d",&n)==1)
{
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 1; i <= n; i++) b[i] = a[i];
map<int,int> mp;
sort(b+1, b+n+1);
int m = unique(b+1, b+n+1) - (b+1);
for(int i=1;i<=m;i++) mp[b[i]] = i ;
for(int i = 1; i<=n; i++) a[i] = mp[a[i]] ,fa[i]=i, last[i]=-1;
dp[0] = 0;
map<int,int> tmp;
for(int i = 1; i <= n; i++)
{
if(~last[a[i]]) U(last[a[i]]-1, last[a[i]]);
last[a[i]] = i;
dp[i] = i;
int c = 0;
for(int j = i; j>0 ; j=f(j-1))
{
c++;
if( c*c > dp[i]) break;
dp[i] = min(dp[i], dp[f(j-1)] + c*c );
}
}
cout<<dp[n]<<endl;
}
}
1009 233 Matrix
首先A[0][n] = A[0][n - 1] * 10 + 3
然后A[m] [n] = A[m] [n - 1] + A[m - 1][n - 1] + ... + A[1][n - 1] + A[0][n]
然后A[m] [n] = A[m] [n - 1] + A[m - 1][n - 1] + ... + A[1][n - 1] + (A[0][n - 1] * 10 + 3)
然后A[m] [n] = 3 + 10 * A[0][n - 1] +A[1] [n - 1] + A[2][n - 1] + ... + A[m - 1][n - 1] + A[m][n - 1]
就能看出，A[*][n] 是A[*][n - 1]的线性表示了。
233 Matrix#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#include<vector>
typedef long long ll;
#define prt(k) cout<< #k" = " << k<< " \n"; ///
const ll mod = 10000007;
int n,m;
void add(ll &a, ll b)
{
a=(a+b)%mod;
}
struct Matrix
{
ll a[22][22];
Matrix()
{
memset(a,0,sizeof a);
}
Matrix operator*(Matrix b)
{
Matrix c;
for(int i = 0; i < n; i++) for(int j =0; j<n; j++)
{
c.a[i][j]= 0;
for(int k = 0; k < n; k ++)
add(c.a[i][j], a[i][k] * b.a[k][j] % mod);
}
return c;
}
void out()
{
for(int i =0; i<n; i++)
{
for(int j = 0 ; j < n; j++) cout<<a[i][j]<<' ';
cout << endl;
}
}
};
Matrix pow(Matrix a, int m)
{
Matrix res; int i,j;
for(i = 0; i < ::n; i++) for(j = 0; j < ::n; j++) res.a[i][j] = (i == j);
for(; m>0; m>>=1, a=a*a) if(m&1) res = res * a;
return res;
}
int main()
{
while(cin >> n >> m)
{
n += 2;
int a[22];
a[0] = 3;
a[1] = 23;
for(int i = 2; i < n; i ++) cin>>a[i];
Matrix ma;
for(int i = 0 ; i < n; i++)
{
int j;
for( j = 0 ; j<=i; j++)
if(j == 1) ma.a[i][j] = 10;
else ma.a[i][j] = 1;
for(; j<n; j++) ma.a[i][j] = 0;
}
;
ma = pow(ma, m);
ll ans= 0; // cout<<ma.a[0][0];
for(int i = 0; i<n; i++) add(ans, ma.a[n-1][i] * a[i] % mod);
cout << ans << endl;
}
}
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1011 Ellipsoid
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define prt(k) cout<<#k" = "<<k<<endl;
const double eps = 1e-8;
int dx[]={0,0,1,1,1,-1,-1,-1};
int dy[]={1,-1,0,1,-1,0,1,-1};
const double inf = 1e30;
double dist(double x,double y,double z)
{
return sqrt(x*x+y*y+z*z);
}
double a,b,c,d,e,f;
double getz(double x,double y)
{
double A=c,B=d*y+e*x,C=a*x*x+b*y*y+f*x*y-1;
double delta=B*B - 4*A*C;
if(delta<0) return inf;
double z1=(-B+sqrt(delta))/2/A;
double z2=(-B-sqrt(delta))/2/A;
if(z1*z1<z2*z2) return z1;
else return z2;
}
double solve()
{
double step=1;
double x=0,y=0,z;
while(step>eps)
{
z=getz(x,y);
for(int i=0;i<8;i++)
{
double nx=x+dx[i]*step;
double ny=y+dy[i]*step;
double nz=getz(nx,ny);
if(nz>=inf) continue;
if(dist(nx,ny,nz)<dist(x,y,z))
{
x=nx,y=ny,z=nz;
}
}
step*=0.99;
}
return dist(x,y,z);
}
int main()
{
while(cin>>a>>b>>c>>d>>e>>f)
{
printf("%.9f\n", solve());
}
return 0;
}