UOJ Round #15 [构造 | 计数 | 异或哈希 kmp]-白红宇

UOJ Round #15 [构造 | 计数 | 异或哈希 kmp]

阅读量：5255 次

发布时间：2019-06-14

本文共 11900 字，大约阅读时间需要 39 分钟。

大部分题目没有AC，我只是水一下部分分的题解...

225

题意：在n*m的棋盘上构造k子棋的平局

题解：

玩一下发现k=1, k=2无解，然后间隔着，上下两行相同：
```
010101010101101010101010
```
这样构造下来就行了。

然后要特判n=1 或 m=1，这时候k=2可以有解

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;typedef long long ll;const int N = 505;inline int read() {    char c=getchar(); int x=0,f=1;    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}    return x*f;}int n, m, k, a[N][N], t[N][N];void paint(int *a, int f) {for(int j=1; j<=m; j++) a[j] = (j&1) ^ f;}inline void _walk(int &x, int &y) {    y++;    if(y == m+1) x++, y=1;}void walk(int &x, int &y, int f) {    while(a[x][y] != f) _walk(x, y);}void solve1() {    if(k == 1) {puts("-1"); return;}    if(n == 1) for(int i=1; i<=m; i++) printf("%d %d\n", 1, i);    else for(int i=1; i<=n; i++) printf("%d %d\n", i, 1);}   int main() {    freopen("in", "r", stdin);    //freopen("out", "w", stdout);    n = read(); m = read(); k = read();    if(min(n, m) == 1) {solve1(); return 0;}    if(k == 1 || k == 2) {puts("-1"); return 0;}    int flag = 0;    if((m&1) && (~n&1)) swap(n, m), flag = 1;    int lim = (n>>1) + 1;    for(int i=1; i <= lim; i++) paint(a[(i<<1)-1], i&1), paint(a[i<<1], i&1);    if(flag) {        for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) t[j][i] = a[i][j];        swap(n, m);        for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) a[i][j] = t[i][j];    }        int b = 0, w = 0;    for(int i=1; i<=n; i++)        for(int j=1; j<=n; j++) {            if(a[i][j] == 1) b++;            else w++;        }    if(b < w) b = 0, w = 1; else b = 1, w = 0;        int bx = 1, by = 1, wx = 1, wy = 1;        lim = n * m;    for(int i=1; i <= lim; i++) {        if(i & 1) walk(bx, by, b), printf("%d %d\n", bx, by), _walk(bx, by);        else walk(wx, wy, w), printf("%d %d\n", wx, wy), _walk(wx, wy);    }        //puts("\ntest");    //for(int i=1; i<=n; i++) for(int j=1; j<=m; j++) printf("%d%c", a[i][j], j==m ? '\n' : ' ');}

226

题意：有向树 / 环 / 基环树的欧拉回路计数

题解：题解太神了看不懂啊，我只能看懂树的...

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;typedef long long ll;const int N = 1e5+5, P = 998244353;inline int read() {    char c=getchar(); int x=0,f=1;    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}    return x*f;}int n, m, u, v, w, de[N], t[N];struct edge {int v, ne;} e[N<<1];int cnt, h[N];inline void ins(int u, int v, int w) {    e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;    e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;}ll inv[N*10], fac[N*10], facInv[N*10];inline ll C(int n, int m) {return fac[n] * facInv[m] %P * facInv[n-m] %P;}int main() {    freopen("in", "r", stdin);    n = read(); m = read();    int flag = 1, lim = 0;    for(int i=1; i<=m; i++) {        u = read(), v = read(), w = t[i] = read(), de[u] += w, de[v] += w;        lim = max(lim, w);        if(w & 1) flag = 0;    }    for(int i=1; i<=n; i++) {        if(de[i] & 1) {puts("0"); return 0;}        de[i] >>= 1;        lim = max(lim, de[i]);    }        inv[1] = fac[0] = facInv[0] = 1;    for(int i=1; i <= lim; i++) {        if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;        fac[i] = fac[i-1] * i %P;        facInv[i] = facInv[i-1] * inv[i] %P;    }    if(m == n-1) {        if(!flag) {puts("0"); return 0;}            ll ans = 1;        for(int i=1; i<=n; i++) ans = ans * fac[de[i] - 1] %P; //printf("ans %lld\n", ans);        for(int i=1; i<=m; i++) ans = ans * C(t[i], t[i] >> 1) %P * (t[i] >> 1) %P;                printf("%lld\n", ans);    }}

227

题意：长为n的序列a，Q次询问，给出一个长m的序列b，求a有多少长m的子序列和b的相对大小关系相同

题解：

太神了不会写啊，看了题解之后写了一下午一晚上才有50部分分

一开始想的是把a的子序列离散化然后和b判断相等

这样对于\(m_i \le 25\)的点可以哈希一下，预处理\(O(nm^2)\)，结果T了

发现题解里的这些点的预处理是\(O(nm\log n)\)的！

想了一下，发现不需要离散化，可以定义每个点的名次为前面比它小的个数+1，照样可行！

然后用树状数组+异或哈希就可以拿这20分啦

试着写了一下AC自动机+主席树的离线做法，还是有细节不会处理啊，又去写了kmp的做法。

其实就是匹配b的名次序列在a中出现次数，只不过a中元素的名次依赖于序列的起始位置。

kmp的border是单调递减的，所以用bit维护当前的border，移动now时暴力修改，求名次用bit求

好喵啊！

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          #include 
           
            double tim() {return (double) clock() / CLOCKS_PER_SEC;}using namespace std;typedef long long ll;const int N = 1e5+5;inline int read() { char c=getchar(); int x=0,f=1; while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();} return x*f;}int type, n, Q, a[N], b[N * 5];void decode(int *b, int m, int ans) { static int c[N]; for(int i=1; i<=m; i++) c[(i+ans-1) % m + 1] = b[i]; for(int i=1; i<=m; i++) b[i] = c[i];} namespace bit { int c[N], t[N], T; inline void add(int p, int v) { for(; p <= n; p += p&-p) { if(t[p] != T) t[p] = T, c[p] = v; else c[p] += v; } } inline int sum(int p) { int ans = 0; for(; p; p -= p&-p) if(t[p] == T) ans += c[p]; return ans; } }namespace m_25 { int s[26][N]; int h[N]; using bit::add; using bit::sum; void ini_hash() { srand(317); for(int i=1; i<=25; i++) h[i] = rand() << 15 ^ rand(); for(int i=1; i<=n; i++) { int ans = 0; bit::T++; for(int m = 1; m <= 25 && i+m-1 <= n; m++) ans = ans << 1 ^ h[sum(a[i+m-1])], s[m][++s[m][0]] = ans, add(a[i+m-1], 1); } for(int m = 1; m <= 25; m++) sort(s[m] + 1, s[m] + s[m][0] + 1); } int get_hash(int *b, int m) { int ans = 0; bit::T++; for(int i=1; i<=m; i++) ans = ans << 1 ^ h[sum(b[i])], add(b[i], 1); return ans; } int Count(int *a, int x) { return upper_bound(a + 1, a + a[0] + 1, x) - lower_bound(a + 1, a + a[0] + 1, x); } int ans; void solve() { ini_hash(); for(int i=1; i<=Q; i++) { int m = read(); for(int i=1; i<=m; i++) b[i] = read(); if(type) decode(b, m, ans); int x = get_hash(b, m); ans = Count(s[m], x); printf("%d\n", ans); } }}namespace kmp { int fail[N], rnk[N], _a[N]; void build(int *s, int n) { bit::T++; for(int i=1; i<=n; i++) rnk[i] = bit::sum(s[i]) + 1, bit::add(s[i], 1); bit::T++; fail[1] = 0; for(int i=2; i<=n; i++) { int now = fail[i-1]; //printf("iiii %d %d\n", i, now); while(now && bit::sum(s[i]) + 1 != rnk[now+1]) { for(int j = i-now; j <= i-fail[now]-1; j++) bit::add(s[j], -1); now = fail[now]; } if(rnk[now+1] == bit::sum(s[i]) + 1) fail[i] = now+1, bit::add(s[i], 1); else fail[i] = 0; } //for(int i=1; i<=n; i++) printf("%d ", rnk[i]); puts(" rnk"); //for(int i=1; i<=n; i++) printf("%d ", fail[i]); puts(" fail"); } void walk(int *a, int n, int *b, int m) { bit::T++; int now = 0, ans = 0; for(int i=1; i<=n; i++) { //printf("-----i %d %d\n", i, now); while(now && bit::sum(a[i]) + 1 != rnk[now+1]) { //printf("lose [%d, %d]\n", i-now, i-fail[now]-1); for(int j = i-now; j <= i-fail[now]-1; j++) bit::add(a[j], -1); now = fail[now]; } if(bit::sum(a[i]) + 1 == rnk[now+1]) now++, bit::add(a[i], 1); if(now == m) ans++; } printf("%d\n", ans); }}namespace type_0 { void solve() { for(int i=1; i<=Q; i++) { int m = read(); for(int j=1; j<=m; j++) b[j] = read(); kmp::build(b, m); kmp::walk(a, n, b, m); for(int i=1; i<=m; i++) kmp::fail[i] = kmp::rnk[i] = 0; } }}int main() { freopen("in", "r", stdin); freopen("out", "w", stdout); type = read(); n = read(); Q = read(); for(int i=1; i<=n; i++) a[i] = read(); if(type == 0) type_0::solve(); else m_25::solve(); return 0;}

下面这个不会写的代码只是留作纪念

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          #include 
           
            double tim() {return (double) clock() / CLOCKS_PER_SEC;}using namespace std;typedef long long ll;const int N = 1e5+5;inline int read() { char c=getchar(); int x=0,f=1; while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();} return x*f;}int type, n, Q, a[N], b[N * 5];void decode(int *b, int m, int ans) { static int c[N]; for(int i=1; i<=m; i++) c[(i+ans-1) % m + 1] = b[i]; for(int i=1; i<=m; i++) b[i] = c[i];} namespace bit { int c[N], t[N], T; inline void add(int p, int v) { for(; p <= n; p += p&-p) { if(t[p] != T) t[p] = T, c[p] = v; else c[p] += v; } } inline int sum(int p) { int ans = 0; for(; p; p -= p&-p) if(t[p] == T) ans += c[p]; return ans; } }namespace m_25 { int s[26][N]; int h[N]; using bit::add; using bit::sum; void ini_hash() { srand(317); for(int i=1; i<=25; i++) h[i] = rand() << 15 ^ rand(); for(int i=1; i<=n; i++) { int ans = 0; bit::T++; for(int m = 1; m <= 25 && i+m-1 <= n; m++) ans = ans << 1 ^ h[sum(a[i+m-1])], s[m][++s[m][0]] = ans, add(a[i+m-1], 1); } for(int m = 1; m <= 25; m++) sort(s[m] + 1, s[m] + s[m][0] + 1); } int get_hash(int *b, int m) { int ans = 0; bit::T++; for(int i=1; i<=m; i++) ans = ans << 1 ^ h[sum(b[i])], add(b[i], 1); return ans; } int Count(int *a, int x) { return upper_bound(a + 1, a + a[0] + 1, x) - lower_bound(a + 1, a + a[0] + 1, x); } int ans; void solve() { ini_hash(); for(int i=1; i<=Q; i++) { int m = read(); for(int i=1; i<=m; i++) b[i] = read(); if(type) decode(b, m, ans); int x = get_hash(b, m); ans = Count(s[m], x); printf("%d\n", ans); } }}namespace tr { #define lc(x) t[x].l #define rc(x) t[x].r struct meow {int l, r, size;} t[N]; int sz, root[N]; void insert(int &x, int l, int r, int p) { t[++sz] = t[x]; x = sz; t[x].size ++; if(l == r) return; int mid = (l+r) >> 1; if(p <= mid) insert(lc(x), l, mid, p); else insert(rc(x), mid+1, r, p); } int sum(int x, int y, int l, int r, int ql, int qr) { //if(ql > qr) return 0; if(ql <= l && r <= qr) return t[y].size - t[x].size; else { int mid = (l+r) >> 1, ans = 0; if(ql <= mid) ans += sum(lc(x), lc(y), l, mid, ql, qr); if(mid < qr) ans += sum(rc(x), rc(y), mid+1, r, ql, qr); return ans; } }}using tr::root; inline int rank(int x, int l, int r) {return tr::sum(root[l-1], root[r], 1, n, 1, a[x]);}namespace ac { #define fir first #define sec second int pos[N]; struct meow { map
            
              ch; map
             
               mp; int fail, cnt, rnk; int count(int c) {return ch.count(c);} int has(int x) {return mp.count(x);} } t[N * 5]; map
              
               ::iterator it; int sz, deep[N]; void insert(int *s, int m, int id) { printf("\ninsert %d\n", id); int u = 0; bit::T++; for(int i=1; i<=m; i++) { int c = s[i]; if(!t[u].ch[c]) { t[u].ch[c] = ++sz; deep[sz] = deep[u] + 1; t[sz].rnk = bit::sum(s[i]) + 1; t[u].mp[t[sz].rnk] = sz; } bit::add(s[i], 1); u = t[u].ch[c]; printf("u %d %d %d\n", u, deep[u], t[u].rnk); } pos[id] = u; } void build() { static int q[N], head = 1, tail = 1; for(it = t[0].ch.begin(); it != t[0].ch.end(); it++) q[tail++] = it -> sec; while(head != tail) { int u = q[head++]; for(it = t[u].ch.begin(); it != t[u].ch.end(); it++) { int v = it -> sec, rnk = t[v].rnk, now = t[u].fail; while(now && !t[now].has(rnk)) now = t[now].fail; if(t[now].has(rnk)) t[v].fail = t[now].mp[rnk]; q[tail++] = v; } } } void walk(int *a, int n) { puts("\nwalk"); int u = 0; for(int i=1; i<=n; i++) { int l = i - deep[u], r = i, rnk = rank(i, l, r); while(u && !t[u].has(rnk)) u = t[u].fail; if(t[u].has(rnk)) u = t[u].mp[rnk]; t[u].cnt ++; printf("u %d\n", u); } } int f[N * 5]; void dfs(int u) { //printf("dfs %d\n", u); f[u] = t[u].cnt; for(map
               
                ::iterator it = t[u].ch.begin(); it != t[u].ch.end(); it++) dfs(it -> sec), f[u] += f[it -> sec]; }}namespace type_0 { void solve() { for(int i=1; i<=n; i++) root[i] = root[i-1], tr::insert(root[i], 1, n, a[i]); //puts("rank"); //for(int l=1; l<=n; l++) for(int r=l; r<=n; r++) printf("rank [%d, %d] %d\n", l, r, rank(r, l, r)); for(int i=1; i<=Q; i++) { int m = read(); for(int j=1; j<=m; j++) b[j] = read(); ac::insert(b, m, i); } ac::build(); ac::walk(a, n); ac::dfs(0); //puts("hi"); using ac::f; using ac::pos; for(int i=1; i<=Q; i++) printf("%d\n", f[pos[i]]); }}int main() { freopen("in", "r", stdin); //freopen("out", "w", stdout); type = read(); n = read(); Q = read(); for(int i=1; i<=n; i++) a[i] = read(); if(type == 0) type_0::solve(); else m_25::solve(); return 0;}