【AtCoder】AGC033(A-F)

AGC033

A - Darker and Darker

直接BFSnode

#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;
char s[1005][1005];
int dis[1005][1005];
bool vis[1005][1005];
queue<pii > Q;
int dx[] = {0,-1,0,1};
int dy[] = {1,0,-1,0};
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) {
        scanf("%s",s[i] + 1);
        for(int j = 1 ; j <= M ; ++j) {
            if(s[i][j] == '#') {
                Q.push(mp(i,j));
                vis[i][j] = 1;
            }
        }
    }
    int ans = 0;
    while(!Q.empty()) {
        pii u = Q.front();Q.pop();
        ans = max(dis[u.fi][u.se],ans);
        for(int k = 0 ; k < 4 ; ++k) {
            int tx = u.fi + dx[k],ty = u.se + dy[k];
            if(tx >= 1 && tx <= N && ty >= 1 && ty <= M) {
                if(!vis[tx][ty]) {
                    vis[tx][ty] = 1;
                    dis[tx][ty] = dis[u.fi][u.se] + 1;
                    Q.push(mp(tx,ty));
                }
            }
        }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

B - LRUD Game

我15min切了第一題c++

第二題想了兩個點函數

首先是分紅兩維ui

\(L,R\)爲例,若是我想從最右邊出去,我確定本身不能用Lspa

可是L會對後手的操做有限制,處理出一個後綴表示後綴中先手使用L後手補救L,往左最多能移動幾個code

當我想拼命使用R時,後手若是左移必須在後綴中先手使用L的峯值以前ci

而後分紅四種,寫成一個函數就好………………rem

掉分了,難過get

#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,N;
int sr,sc;
char s[MAXN],t[MAXN];
int suf[MAXN];
bool Process(char a,char b,int dis,int lim) {
    suf[N + 1] = 0;int sum = 0;
    for(int i = N ; i >= 1 ; --i) {
        suf[i] = suf[i + 1];
        if(s[i] == b) {
            suf[i] = min(suf[i],sum - 1);
            --sum;
        }
        if(t[i] == a) {
            if(sum < 0) ++sum;
        }
    }
    int pos = dis;
    for(int i = 1 ; i <= N ; ++i) {
        if(s[i] == a) ++pos;
        if(pos > lim) return false;
        if(t[i] == b) {
            if(pos - 1 + suf[i + 1] >= 1) --pos;
        }
    }
    return true;
}
void Solve() {
    read(H);read(W);read(N);
    read(sr);read(sc);
    scanf("%s",s + 1);scanf("%s",t + 1);
    if(!Process('D','U',sr,H) || !Process('U','D',H + 1 - sr,H) ||
    !Process('R','L',sc,W) || !Process('L','R',W + 1 - sc,W)) {
        puts("NO");
    }
    else puts("YES");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

C - Removing Coins

又是考後兩分鐘想出題解,哭了it

由於B題沒切C題看的心煩意亂

每次至關於割掉全部葉子,若是選的根是一個葉子則保留一個葉子

若是是一條鏈,長度爲1後手勝,長度爲2先手勝,長度爲3先手勝,長度爲4後手勝,是這樣一個循環……

對應到樹上呢,很神奇,是樹的直徑!

由於咱們對於直徑也有一次割兩個或者一次割一個的操做……

就……作完了

#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);
}
struct node {
    int to,next;
}E[MAXN * 2];
int sumE,head[MAXN],N;
int dis[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u,int fa) {
    dis[u] = dis[fa] + 1;
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(v != fa) {
            dfs(v,u);
        }
    }
}
void Solve() {
    read(N);
    if(N == 1) {
        puts("First");return;
    }
    int a,b;
    for(int i = 1 ; i < N ; ++i) {
        read(a);read(b);
        add(a,b);add(b,a);
    }
    dfs(1,0);
    int u = 1;
    for(int i = 1 ; i <= N ; ++i) {
        if(dis[i] > dis[u]) u = i;
    }
    dfs(u,0);
    u = 1;
    for(int i = 1 ; i <= N ; ++i) {
        if(dis[i] > dis[u]) u = i;
    }
    if(dis[u] % 3 == 2) {puts("Second");}
    else puts("First");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Complexity

答案確定不會超過\(log(HW)\),最可能是15

記錄計算\((r,c)\)\((r' ,c)\)這兩個點最右能延伸到哪一列

以及\((r,c)\)\((r,c')\)最下能延伸到哪一行

橫豎能夠交替轉移

狀態壓縮一下空間就是\(n^3\)

複雜度的話多是\(n^4\)可是有個小常數

#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 205
//#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,sum[MAXN][MAXN];
int f[2][MAXN][MAXN][MAXN],g[2][MAXN][MAXN][MAXN];
char s[MAXN][MAXN];
bool check(int xa,int ya,int xb,int yb) {
    if(yb > W || xb > H) return false;
    int all = (yb - ya + 1) * (xb - xa + 1);
    int t = sum[xb][yb] + sum[xa - 1][ya - 1] - sum[xb][ya - 1] - sum[xa - 1][yb];
    if(t == all || t == 0) return true;
    return false;
}
void Solve() {
    read(H);read(W);
    for(int i = 1 ; i <= H ; ++i) {
    scanf("%s",s[i] + 1);
    for(int j = 1 ; j <= W ; ++j) sum[i][j] = (s[i][j] == '#');
    for(int j = 1 ; j <= W ; ++j) sum[i][j] += sum[i][j - 1];
    }
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) sum[i][j] += sum[i - 1][j];
    }
    int cur = 0;
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = i ; j <= H ; ++j) {
        for(int c = 1 ; c <= W ; ++c) {
        int t = f[cur][i][j][c - 1];
        if(check(i,c,j,c)) t = max(t,c);
        if(t >= c) {
            while(check(i,c,j,t + 1)) ++t;
        }
        else t = c - 1;
        f[cur][i][j][c] = t;
        }
    }
    }
    for(int i = 1 ; i <= W ; ++i) {
    for(int j = i ; j <= W ; ++j) {
        for(int r = 1 ; r <= H ; ++r) {
        int t = g[cur][i][j][r - 1];
        if(check(r,i,r,j)) t = max(t,r);
        if(t >= r) {
            while(check(r,i,t + 1,j)) ++t;
        }
        else t = r - 1;
        g[cur][i][j][r] = t;
        }
    }
    }
    for(int k = 0 ; k <= 20 ; ++k) {
    if(f[cur][1][H][1] >= W || g[cur][1][W][1] >= H) {
        out(k);enter;return;
    }
    memset(f[cur ^ 1],0,sizeof(f[cur ^ 1]));
    memset(g[cur ^ 1],0,sizeof(g[cur ^ 1]));
    for(int i = 1 ; i <= H ; ++i) {
        for(int j = i ; j <= H ; ++j) {
        for(int c = 1 ; c <= W ; ++c) {
            int t = f[cur][i][j][c];
            if(t >= W) {f[cur ^ 1][i][j][c] = W;continue;}
            t = f[cur][i][j][t + 1];
            while(1) {
            int k = g[cur][c][t + 1][i];
            k = g[cur][c][t + 1][k + 1];
            if(k >= j) ++t;
            else break;
            }
            f[cur ^ 1][i][j][c] = t;
        }
        }
    }
    for(int i = 1 ; i <= W ; ++i) {
        for(int j = i ; j <= W ; ++j) {
        for(int r = 1 ; r <= H ; ++r) {
            int t = g[cur][i][j][r];
            if(t >= H) {g[cur ^ 1][i][j][r] = H;continue;}
            t = g[cur][i][j][t + 1];
            while(1) {
            int k = f[cur][r][t + 1][i];
            k = f[cur][r][t + 1][k + 1];
            if(k >= j) ++t;
            else break;
            }
            g[cur ^ 1][i][j][r] = t;
        }
        }
    }
    cur ^= 1;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Go around a Circle

……E題咋也那麼簡單

失去漲分良機

這個就是,不妨設第一次出現的字母是R,那麼B確定不會在環上連續出現兩次

咱們斷開B,就造成了一個個小島

小島的長度必須是奇數,由於若是是偶數則中心點走到兩邊只有奇數,而端點到端點就只有偶數,而第一段連續R不可能又是奇數又是偶數

若是第一段連續的R長度是t,那麼小島的長度最大是t+1,不然小島的長度最大是t

而中間的B呢,至關於咱們更換小島,或者呆在原來的小島上,可是因爲每一個點都會經歷中間的B,因此就算換了小島,每一個小島的兩邊都會有點

咱們要走偶數個R,只須要來回走一條邊

而奇數個點,咱們要走到小島的另外一邊,這個就對小島長度有了限制

注意最後一段R不限制咱們的小島

這樣的話就能夠dp了

特殊狀況是若是全是R那麼小島的長度能夠無窮大

(我爲啥要起小島這樣的名字呢2333)

#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);
}
const int MOD = 1000000007;
int N,M,L;
char s[MAXN];
bool flag = 0;
int f[MAXN][2],dp[MAXN],sum[MAXN][2];
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);
}
void Solve1() {
    int ans = 0;
    f[1][1] = 1;f[1][0] = 1;
    for(int i = tijie2 ; i <= N ; ++i) {
    f[i][0] = f[i - 1][1];
    f[i][1] = inc(f[i - 1][1],f[i - 1][0]);
    }
    ans = inc(f[N][0],f[N][1]);
    memset(f,0,sizeof(f));
    f[1][0] = 1;
    for(int i = 2 ; i < N ; ++i) {
    f[i][0] = f[i - 1][1];
    f[i][1] = inc(f[i - 1][1],f[i - 1][0]);
    }
    ans = inc(ans,MOD - f[N - 1][1]);
    out(ans);enter;
}
void Solve2() {
    dp[0] = 1;sum[0][0] = 1;
    for(int i = 1 ; i <= N ; ++i) {
    if(i >= 2) {
        int t = sum[i - 2][i & 1];
        if(i > L + 1) update(t,MOD - sum[i - L - 2][i & 1]);
        dp[i] = t;
    }
    sum[i][0] = sum[i - 1][0];sum[i][1] = sum[i - 1][1];
    sum[i][i & 1] = inc(sum[i][i & 1],dp[i]);
    
    }
    int ans = dp[N];
    for(int i = 1 ; i <= L ; i += 2) {
    update(ans,mul(i,dp[N - i - 1]));
    }
    out(ans);enter;
}
void Init() {
    read(N);read(M);
    scanf("%s",s + 1);
    while(M >= 1 && s[M] == s[1]) --M;
    L = N;
    if(M == 0) flag = 1;
    else {
    for(int i = 1 ; i <= M ; ++i) {
        if(s[i] != s[1]) continue;
        else {
        int j = i;
        while(s[j + 1] == s[1]) ++j;
        int t = j - i + 1;
        if(i == 1) {
            if(t % 2 == 0) L = min(L,t + 1);
            else L = min(L,t); 
        }
        else {
            if(t % 2 == 1) L = min(L,t);
        }
        i = j;
        }
    }
    }
    if(L == N) flag = 1;
    if(flag) Solve1();
    else Solve2();
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
}

F - Adding Edges

彷佛就是個暴力?

咱們只要每次把存在一條邊\((a,b)\),c在\((a,b)\)的路徑上,且存在\((a,c)\),刪掉\((a,b)\),加上\((b,c)\),最後只須要對於每一個點當作根,在這些路徑上走從祖先到兒子的邊,不然不走,每一個點搜出來的點的個數(除了本身)就是這個點能夠連出去的邊的總數

怎麼搜,就是直接加一條邊的時候,找出這條鏈,而後找這條鏈兩端看能不能縮小一個範圍,而後新加的邊和兩個端點以前的邊比較一下,能刪則刪

#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 2005
//#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);
}
struct node {
    int to,next;
}E[MAXN * 2];
int N,M,sumE,head[MAXN],dis[MAXN][MAXN];
int fa[MAXN][MAXN];
bool vis[MAXN][MAXN];
bool g[MAXN][MAXN];
set<int> S[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int id,int u) {
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(v != fa[id][u]) {
            fa[id][v] = u;
            dis[id][v] = dis[id][u] + 1;
            dfs(id,v);
        }
    }
}
vector<int> getline(int a,int b) {
    vector<int> res;
    while(a != b) {
        res.pb(a);
        a = fa[b][a];
    }
    res.pb(b);
    return res;
}
void build(int a,int b) {
    vis[a][b] = vis[b][a] = 1;
    S[a].insert(b);S[b].insert(a);
}
void destroy(int a,int b) {
    vis[a][b] = vis[b][a] = 0;
    S[a].erase(b);S[b].erase(a);
}
void AddE(int a,int b) {
    if(vis[a][b]) return;
    vector<int> v = getline(a,b);
    int s = v.size() - 1;
    int f = a,t = b,q = 0;
    for(int i = 0 ; i < s ; ++i) {
        if(vis[f][v[i]]) {f = v[i];q = i;}
    }
    for(int i = s - 1 ; i >= q ; --i) {
        if(vis[t][v[i]]) t = v[i];
    }
    if(t == a) return;
    v.clear();
    if(vis[f][t]) return;

    vector<pii> E;
    for(auto k : S[f]) {
        if(dis[f][k] == dis[f][t] + dis[t][k]) {
            E.pb(mp(t,k));
            v.pb(k);
        }
    }
    for(auto k : v) destroy(k,f);
    v.clear();
    for(auto k : S[t]) {
        if(dis[t][k] == dis[t][f] + dis[f][k]) {
            E.pb(mp(f,k));
            v.pb(k);
        }
    }
    for(auto k : v) destroy(k,t);
    v.clear();
    build(f,t);
    for(auto k : E) {
        AddE(k.fi,k.se);
    }
}
void find(int id,int u) {
    g[id][u] = 1;
    for(auto t : S[u]) {
        if(g[id][t]) continue;
        if(dis[t][id] == dis[id][u] + dis[u][t]) find(id,t);
    }
}
void Solve() {
    read(N);read(M);
    int a,b;
    for(int i = 1 ; i < N ; ++i) {
        read(a);read(b);
        add(a,b);add(b,a);
    }
    for(int i = 1 ; i <= N ; ++i) dfs(i,i);
    for(int i = 1 ; i <= M ; ++i) {
        read(a);read(b);
        AddE(a,b);
    }
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
        find(i,i);
        for(int j = i + 1 ; j <= N ; ++j) if(g[i][j]) ++ans;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

明天期中了,rp++
(但願別墊底,枯了)

ps:我發現yutaka的題故意會把B題放的難一點。。。

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