AtCoderAGC016

Posted ivorysi

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A - Shrinking

用每个字母模拟一下就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
char s[105];
bool vis[30],a[105],b[105];
int N;

void Solve() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
    for(int i = 1 ; i <= N ; ++i) vis[s[i] - 'a'] = 1;
    int ans = N;
    for(int t = 0 ; t < 26 ; ++t) {
    if(vis[t]) {
        int cnt = 0;
        memset(a,0,sizeof(a));
        for(int i = 1 ; i <= N ; ++i) a[i] = (s[i] == 'a' + t);
        while(1) {
        bool f = 1;
        for(int i = 1 ; i <= N - cnt; ++i) {
            f = f & a[i];
        }
        if(f) break;
        memset(b,0,sizeof(b));
        for(int i = 1 ; i <= N - cnt - 1; ++i) {
            b[i] = a[i] || a[i + 1];
        }
        ++cnt;
        memcpy(a,b,sizeof(a));
        
        }
        ans = min(ans,cnt);
    }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

B - Colorful Hats

如果最大值和最小值相等，要么最大值等于N - 1，否则就是最大值乘2小于N
如果相差1，最小值的个数是t
那么要满足最大值\(t + 1\leq A \leq t + (N - t) / 2\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,a[MAXN];
int tot;
void Solve() {
    read(N);
    int maxx = 0,minn = N;
    for(int i = 1 ; i <= N ; ++i) {
    read(a[i]);maxx = max(a[i],maxx);minn = min(a[i],minn);
    }
    if(maxx - minn > 1) {puts("No");return;}
    if(maxx == minn) {
    if(maxx == N - 1) puts("Yes");
    else if(maxx * 2 <= N)  puts("Yes");
    else puts("No");
    return;
    }
    
    for(int i = 1 ; i <= N ; ++i) {
    if(a[i] == minn) ++tot;
    }
    if(maxx <= tot) {puts("No");return;}
    if(N - tot >= 2 * (maxx - tot)) {
    puts("Yes");return;
    }
    puts("No");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

C - +/- Rectangle

如果\(H\)是\(h\)的倍数并且\(W\)是\(w\)的倍数，那么无解

否则认为\(H\)不是\(h\)的倍数，以0开始标号，每个\(h\)的倍数的行都填成正数\(1000(h - 1) - 1\)，其他行都填成-1000

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int H,W,h,w;
void Solve() {
    read(H);read(W);read(h);read(w);
    if(H % h != 0) {
    puts("Yes");
    for(int i = 0 ; i < H ; ++i) {
        for(int j = 0 ; j < W ; ++j) {
        if(i % h == 0) {out(1000 * (h - 1) - 1);}
        else out(-1000);
        space;
        }
        enter;
    }
    }
    else if(W % w != 0) {
    puts("Yes");
    for(int i = 0 ; i < H ; ++i) {
        for(int j = 0 ; j < W ; ++j) {
        if(j % w == 0) {out(1000 * (w - 1) - 1);}
        else out(-1000);
        space;
        }
        enter;
    }
    }
    else {
    puts("No");
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - XOR Replace

相当于最后多了一个位置，是a的异或和，然后每次相当于交换最后多的这个位置上的数和选中的数

然后我们按照权值建点，a和b对应位置的数值连边，起点为a的异或和，终点为多出来的数，这两点中间连一条边，我们需要每个联通块有欧拉回路，所以统计奇数点的个数，除2是附加边的个数，然后走完一个联通块到下一个联通块中不用走回来，所以是多连一条即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
int a[MAXN],b[MAXN],N,val[MAXN],tot,st,ed,deg[MAXN],cnt;
int fa[MAXN];
map<int,int> zz;
int getfa(int u) {
    return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
void Init() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);a[N + 1] ^= a[i];}
    for(int i = 1 ; i <= N ; ++i) read(b[i]);
    for(int i = 1 ; i <= N + 1 ; ++i) {
        zz[a[i]]++;
    }
}
void Solve() {
    for(int i = 1 ; i <= N ; ++i) {
        if(zz[b[i]] == 0) {
            puts("-1");return;
        }
        zz[b[i]] -= 1;
    }
    st = a[N + 1];
    for(int i = 1 ; i <= N + 1; ++i) {
        val[++tot] = a[i];
        if(zz[a[i]]) ed = a[i];
    }
    sort(val + 1,val + tot + 1);
    tot = unique(val + 1,val + tot + 1) - val - 1;
    st = lower_bound(val + 1,val + tot + 1,st) - val;
    ed = lower_bound(val + 1,val + tot + 1,ed) - val;
    for(int i = 1 ; i <= tot ; ++i) {
        fa[i] = i;
    }
    for(int i = 1 ; i <= N ; ++i) {
        if(a[i] != b[i]) {
            int p = lower_bound(val + 1,val + tot + 1,a[i]) - val;
            int q = lower_bound(val + 1,val + tot + 1,b[i]) - val;
            ++deg[p];++deg[q];
            fa[getfa(p)] = getfa(q);
            ++cnt;
        }
    }
    ++deg[st];++deg[ed];
    int p = 0;
    for(int i = 1 ; i <= tot ; ++i) {
        if(deg[i] & 1) ++p;
    }
    cnt += p / 2;
    for(int i = 1 ; i <= tot ; ++i) {
        if(deg[i]) {
            if(getfa(i) != getfa(st)) {
                fa[getfa(i)] = getfa(st);
                cnt++;
            }
        }
    }
    out(cnt);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}

E - Poor Turkeys

如果一个点集S在t次操作后不会被消灭，有四种情况

\(x_{t} \in S,y_{t} \in S\)那么一定会被消灭

如果\(x_{t} \in S,y_{t} \notin S\)那么前\(t - 1\)次\(S \cup y_{t}\)必须得存在

如果\(x_{t} \notin S,y_{t} \in S\)那么前\(t - 1\)次\(S\cup x_{t}\)必须得存在

如果\(x_{t} \notin S,y_{t} \notin S\)那么前\(t - 1\)次\(S\)必须存在

所以对于一个点\(v\)是否最后可以存在，那么可以用\({v}\)往后倒推

对于一个点对必须存在

\(v\)不会被吃掉

\(u\)不会被吃掉

操作两个点对得到的集合没有相同的点

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M;
int x[MAXN],y[MAXN];
bool S[405][405],eat[405];
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= M ; ++i) {
        read(x[i]);read(y[i]);
    }
    for(int i = 1 ; i <= N ; ++i) {
        S[i][i] = 1;
        for(int j = M ; j >= 1 ; --j) {
            if(S[i][x[j]] && S[i][y[j]]) {eat[i] = 1;break;}
            else if(S[i][x[j]] && !S[i][y[j]]) S[i][y[j]] = 1;
            else if(!S[i][x[j]] && S[i][y[j]]) S[i][x[j]] = 1;
        }
    }
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = i + 1 ; j <= N ; ++j) {
            if(!eat[i] && !eat[j]) {
                bool flag = 1;
                for(int h = 1 ; h <= N ; ++h) {
                    if(S[i][h] && S[j][h]) {flag = 0;break;}
                }
                ans += flag;
            }
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - Games on DAG

用\(dp[S]\)表示给\(S\)之间的点加边使得1和2的sg函数相同

我们可以把\(S\)划分成两个点集\(T\)和\(U\)保证这两个点集要么都有12，要么12一个都没有

\(U\)之间没有边

\(U\)到\(T\)的边随意

\(T\)到\(U\)必须有一条边

然后\(T\)之间的边连边方式是\(dp[T]\)

答案是\(2^{M} - dp[2^{N} - 1]\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M;
int c[16],dp[(1 << 15) + 5],cnt[(1 << 15) + 5],pw[100005];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int lowbit(int x) {
    return x & (-x);
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);read(M);
    pw[0] = 1;
    for(int i = 1 ; i <= M ; ++i) pw[i] = mul(pw[i - 1],2);
    for(int i = 1 ; i < (1 << N) ; ++i) cnt[i] = cnt[i - lowbit(i)] + 1;
    int x,y;
    for(int i = 1 ; i <= M ; ++i) {
        read(x);read(y);
        c[x] |= 1 << y - 1;
    }
    dp[0] = 1;
    for(int S = 1 ; S < (1 << N) ; ++S) {
        if((S & 3) != 3 && (S & 3) != 0) continue;
        for(int T = S; T ; T = (T - 1) & S) {
            if((T & 3) != 3 && (T & 3) != 0) continue;
            int d = 1;
            for(int i = 1 ; i <= N ; ++i) {
                if(T & (1 << i - 1)) d = mul(d,pw[cnt[c[i] & (S ^ T)]]);
                if((S ^ T) & (1 << i - 1)) d = mul(d,pw[cnt[c[i] & T]] - 1);
            }
            d = mul(d,dp[S ^ T]);
            update(dp[S],d);
        }
    }
    out(inc(pw[M],MOD - dp[(1 << N) - 1]));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

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