2014 Beijing Invitational competition abdhj

A. a matrix click to open the link

Structure. The conclusion is that a curve is generated starting from the first line, so that the interval is longest and descending from top to bottom,

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <stdio.h>#include <vector>#include <set>using namespace std;#define N 100005vector<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

B: Click the open link.

Violence

#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

E: Click the open link.

Obtain a DP equation first.

DP [I] [J] indicates that the string length is I, and there are different numbers of J methods at the end.

Then we get the transfer equation and use the Matrix to accelerate the power.

#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 20140518struct 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

H: Click the open link.

Sort all the numbers together after obtaining them, and the maximum value is reduced to the minimum.

If the maximum and minimum values are from the same number, the answer is the second largest smallest or the second smallest.

#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: Click to open the link.

Hash: the first location with different characters at 0, and the second location with different characters at the first character.

#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;}