2014 北京邀请赛ABDHJ题解

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A. A Matrix

点击打开链接

构造，结论是从第一行开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 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:点击打开链接

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

E：点击打开链接

先得到一个dp方程

dp[i][j]表示字符串长度为i，结尾有j个数互不相同的方法数

然后得到转移方程再用矩阵快速幂加速

#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 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：点击打开链接

把所有数字取法后都一起排序，最大的减最小的即可

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;}

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