【AtCoder】ARC081

C - Make a Rectangle

每次取兩個相同的且最大的邊，取兩次便可

#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-12
//#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];
map<int,int> zz;
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);zz[a[i]]++;}
    int r = 0,c = 0;
    while(1) {
        if(zz.empty()) break;
        auto t = *(--zz.end());zz.erase(--zz.end());
        if(t.se >= 2) {
            if(!r) r = t.fi;
            else if(!c) c = t.fi;
        }
        if(t.se > 2) zz[t.fi] = t.se - 2;
        if(r && c) break;
    }
    out(1LL * r * c);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Coloring Dominoes

就是若是是
豎條對兩個長條，那麼方案數乘上2
豎條對豎條，乘上2
兩個長條+豎條，方案數乘上1
兩個長條+兩個長條 有3種
1 2
2 1
和
1 3
2 1
和
1 2
2 3

初始若是一個豎條有3種，兩個橫條有6種

#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-12
//#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,f[65];
char s[2][65];
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 Solve() {
    read(N);
    scanf("%s%s",s[0] + 1,s[1] + 1);
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
        if(s[0][i] == s[0][i + 1]) continue;
        if(s[0][i - 1] == s[0][i]) {
            if(i <= 2) ans = 6;
            else {
                if(f[i - 2] == 0) ans = mul(ans,3);
                else if(f[i - 2] == 1) ans = mul(ans,2);
            }
            f[i] = 0;
        }
        else {
            if(i <= 1) ans = 3;
            else {
                if(f[i - 1] == 1) ans = mul(ans,2);
            }
            f[i] = 1;
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Don't Be a Subsequence

撞題了吧，最短不公共子串的那個模板大禮包

直接建序列自動機，求最短路，每次從前日後遍歷走一條最短路邊

#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 200005
#define eps 1e-12
//#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 a[MAXN];
int nxt[MAXN][26],N,dis[MAXN];
string ans = "";
void Solve() {
    scanf("%s",a + 1);
    N = strlen(a + 1);
    for(int i = 0 ; i < 26 ; ++i) nxt[N][i] = N + 1;
    for(int i = N - 1 ; i >= 0 ; --i) {
        for(int j = 0 ; j < 26 ; ++j) nxt[i][j] = nxt[i + 1][j];
        nxt[i][a[i + 1] - 'a'] = i + 1;
    }
    dis[N + 1] = 0;
    for(int i = N ; i >= 0 ; --i) {
        dis[i] = N + 1;
        for(int j = 0 ; j < 26 ; ++j) dis[i] = min(dis[nxt[i][j]] + 1,dis[i]);
    }
    int pos = 0;
    while(pos != N + 1) {
        for(int i = 0 ; i < 26 ; ++i) {
            if(dis[nxt[pos][i]] + 1 == dis[pos]) {
                ans += (i + 'a');
                pos = nxt[pos][i];
                break;
            }
        }
    }
    cout << ans << endl;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Flip and Rectangles

咱們認爲一個長方形任意一個22的方格若是包含偶數個黑點
那麼必定會變成全黑
咱們把每一個22的方格中間標記成一個新點，判斷可否選擇，而後用最大子矩形的算法作

#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 2005
#define eps 1e-12
//#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,ans;
char s[MAXN][MAXN];
int tr[MAXN],tl[MAXN],h[MAXN],l[MAXN],r[MAXN];
bool check(int x,int y) {
    return ((s[x - 1][y - 1] == '#') + (s[x - 1][y] == '#') + (s[x][y - 1] == '#') + (s[x][y] == '#')) & 1;
}
void Solve() {
    read(H);read(W);
    ans = max(H,W);
    for(int i = 0 ; i < H ; ++i) {
        scanf("%s",s[i]);
    }
    for(int i = 1 ; i <= W - 1; ++i) l[i] = 0,r[i] = W;
    tr[W] = W;
    for(int i = 1 ; i < H ; ++i) {
        for(int j = 1 ; j < W ; ++j) {
            if(check(i,j)) tl[j] = j;
            else tl[j] = tl[j - 1];
        }
        for(int j = W - 1 ; j >= 1 ; --j) {
            if(check(i,j)) tr[j] = j;
            else tr[j] = tr[j + 1];
        }
        for(int j = 1 ; j < W ; ++j) {
            if(check(i,j)) {
                h[j] = 0,l[j] = 0,r[j] = W;
            }
            else {
                l[j] = max(tl[j],l[j]);
                r[j] = min(tr[j],r[j]);
                h[j]++;
                ans = max(ans,(r[j] - l[j] - 1 + 1) * (h[j] + 1));
            }
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}