AtCoderARC089

Posted 2021-01-25 ivorysi

C - Traveling

先看能不能走到，再看看奇偶性是否相同

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar(‘
‘)
#define space putchar(‘ ‘)
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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;
void Solve() {
    int pt,px,py;
    int t,x,y;
    read(N);
    pt = 0,px = 0,py = 0;
    for(int i = 1 ; i <= N ; ++i) {
    read(t);read(x);read(y);
    int k = abs(x - px) + abs(y - py);
    if(k > t - pt) {
        puts("No");return;
    }
    if((k ^ t) & 1) {puts("No");return;}
    }
    puts("Yes");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Checker

我们计算右下角在((-2k,-2k))到((-1,-1))这个区域内，每个点所在的格子的颜色

发现根据右下角的位置会分成九个小块，把九个小块里的颜色和需求一样的矩阵用差分矩阵加，最后统计前缀和中最大的即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar(‘
‘)
#define space putchar(‘ ‘)
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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,K;
int a[2005][2005];
int dx[5],dy[5];
void add(int x1,int y1,int x2,int y2) {
    if(x1 > x2 || y1 > y2) return;
    a[x1][y1]++;a[x2 + 1][y2 + 1]++;
    a[x1][y2 + 1]--;a[x2 + 1][y1]--;
}
void Solve() {
    read(N);read(K);
    int x,y;char c[5];
    for(int i = 1 ; i <= N ; ++i) {
    read(x);read(y);scanf("%s",c + 1);
    ++x;++y;
    int tx = x % K,ty = y % K;
    int now = ((x / K) ^ (y / K)) & 1,u;
    if(c[1] == ‘W‘) u = 0;
    else u = 1;
    dx[1] = K - tx,dx[2] = K,dx[3] = tx;
    dy[1] = K - ty,dy[2] = K,dy[3] = ty;
    for(int i = 1 ; i <= 3 ; ++i) dx[i] += dx[i - 1],dy[i] += dy[i - 1];
    for(int h = 1 ; h <= 3 ; ++h) {
        for(int t = 1 ; t <= 3 ; ++t) {
        int w = (h ^ t ^ now) & 1;
        if(w == u) {
            add(dx[h - 1] + 1,dy[t - 1] + 1,dx[h],dy[t]);
        }
        }
    }
    }
    int ans = 0;
    for(int i = 1 ; i <= 2 * K ; ++i) {
    for(int j = 1 ; j <= 2 * K ; ++j) {
        a[i][j] = a[i][j] + a[i][j - 1] + a[i - 1][j] - a[i - 1][j - 1];
        ans = max(ans,a[i][j]);
    }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - GraphXY

显然(ax + by + f(a,b) >= d(x,y))
(a)最多100个(b)最多100个，可以都构建出来

然后对于每个((a,b))求出来(f(a,b))
最后再对于每个(x,y)判一遍(d(x,y))是否合法

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar(‘
‘)
#define space putchar(‘ ‘)
#define eps 1e-8
#define mo 974711
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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,B;
int d[15][15];
int f[105][105],tot,gx[105],gy[105],S,T;
pii p[305 * 305];
int val[305 * 305],cnt;
int check(int x,int y) {
    int res = 1000000;
    for(int i = 0 ; i <= 100 ; ++i) {
    for(int j = 0 ; j <= 100 ; ++j) {
        res = min(res,f[i][j] + i * x + j * y);
    }
    }
    return res;
}
void Solve() {
    read(A);read(B);
    for(int i = 1 ; i <= A ; ++i) {
    for(int j = 1 ; j <= B ; ++j) {
        read(d[i][j]);
    }
    }
    for(int i = 0 ; i <= 100 ; ++i) {
    for(int j = 0 ; j <= 100 ; ++j) {
        for(int k = 1 ; k <= A ; ++k) {
        for(int h = 1 ; h <= B ; ++h) {
            f[i][j] = max(f[i][j],d[k][h] - k * i - h * j);
        }
        }
    } 
    }
    for(int i = 1 ; i <= A ; ++i) {
    for(int j = 1 ; j <= B ; ++j) {
        if(check(i,j) != d[i][j]) {
        puts("Impossible");
        return;
        }
    }
    }
    puts("Possible");
    S = ++tot;
    gx[0] = S;
    for(int i = 1 ; i <= 100 ; ++i) {
    gx[i] = ++tot;
    p[++cnt] = mp(gx[i - 1],gx[i]);
    val[cnt] = -2;
    }
    T = ++tot;
    gy[0] = T;
    for(int i = 1 ; i <= 100 ; ++i) {
    gy[i] = ++tot;
    p[++cnt] = mp(gy[i],gy[i - 1]);
    val[cnt] = -1;
    }
    for(int i = 0 ; i <= 100 ; ++i) {
    for(int j = 0 ; j <= 100 ; ++j) {
        p[++cnt] = mp(gx[i],gy[j]);
        val[cnt] = f[i][j];
    }
    }
    out(tot);space;out(cnt);enter;
    for(int i = 1 ; i <= cnt ; ++i) {
    out(p[i].fi);space;out(p[i].se);space;
    if(val[i] < 0) {
        if(val[i] == -2) {puts("X");}
        else puts("Y");
    }
    else {out(val[i]);enter;}
    }
    out(S);space;out(T);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - ColoringBalls

本来以为是个dp没想到是个搜索加剪枝，迷的很。。然而就是搜索我都写跪了……菜的要死QAQ

我们对于白块分成的每一个小块进行分组
然后把相邻的同颜色合起来，比如说RRBBRR变成RBR
然后我们根据所需要达到的最小操作次数来分组

1次操作 r R
2次操作 rb B BR RB BRB
3次操作 rb? BRB RBRB BRBR RBRBR
4次操作 rb?? BRBRB RBRBRB BRBRBR RBRBRBR

问号表示可以任意一种操作

然后对于一种序列，我们可以把必须要填的位置填上，然后挖上几个坑，往里扔数，方案数就是组合数了
例如序列 2 2 3
我们必须要填的是
BWBWBRB
然后挖的坑是
W/R/B/R/W/R/B/R/W/R/B/R/B/R/W
往坑里扔数就行

怎么判断一个序列合不合法，找出所有的r以及它们最靠左没有被搭配过的b，从后往前枚举r，从小到大枚举所需要操作序列长度，对于一个b加上我后面需要用的位置的个数，统计一个后缀和看看合不合法即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar(‘
‘)
#define space putchar(‘ ‘)
#define eps 1e-8
#define mo 974711
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    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,K,C[1005][1005],L[75],fac[1005],invfac[1005],cnt;
int sum[75],pos[75],tot,ans,matc[75];
char s[75];
bool vis[75];
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;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
bool check() {
    if(tot < cnt) return false;
    memset(sum,0,sizeof(sum));
    int p = 1;
    for(int i = cnt ; i >= 1 ; --i) {
    sum[pos[i]]++;
    if(L[p] >= 2) {
        if(!matc[pos[i]]) return false;
        sum[matc[pos[i]]] += L[p] - 1;
    }
    ++p;
    }
    for(int i = K ; i >= 1 ; --i) {
    sum[i] += sum[i + 1];
    if(sum[i] > K - i + 1) return false;
    }
    return true;
}
bool dfs(int pre,int dep,int len) {
    cnt = dep;
    if(!check()) return false;
    int k = 1 + cnt;
    for(int i = 1 ; i <= cnt ; ++i) {
        if(L[i] == 1) ++k;
        else k += 2 * L[i] - 1;
    }
    int res = C[N - len + k - 1][k - 1];
    res = mul(res,fac[cnt]);
    int t = 0;
    for(int i = 1 ; i <= cnt ; ++i) {
        if(L[i] != L[i - 1]) {
            res = mul(res,invfac[t]);
            t = 0;
        }
        ++t;
    }
    res = mul(res,invfac[t]);
    update(ans,res);
    if(dep + 1 > tot) return true;
    for(int i = pre ; i <= 70 ; ++i) {
        int tl = len;
        if(dep != 0) ++tl;
        if(i == 1 || i == 2) tl += 1;
        else tl += i - 2 + i - 1;
        if(tl > N) break;
        L[dep + 1] = i;
        if(!dfs(i,dep + 1,tl)) break;
    }
    return true;
}
void Solve() {
    read(N);read(K);
    scanf("%s",s + 1);
    C[0][0] = 1;
    for(int i = 1 ; i <= 1000 ; ++i) {
        C[i][0] = 1;
        for(int j = 1 ; j <= i ; ++j) {
            C[i][j] = inc(C[i - 1][j - 1],C[i - 1][j]);
        }
    }
    fac[0] = 1;
    for(int i = 1 ; i <= 1000 ; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[1000] = fpow(fac[1000],MOD - 2);
    for(int i = 999 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    tot = 0;
    for(int i = 1 ; i <= K ; ++i) {
        if(s[i] == ‘r‘) {
        pos[++tot] = i;
    
        for(int j = i + 1 ; j <= K ; ++j) {
        if(s[j] == ‘b‘ && !vis[j]) {matc[i] = j;vis[j] = 1;break;}
        }
    }
    }
    dfs(1,0,0);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}