【AtCoder】AGC004

時間  2019-12-04
標籤 AtCoder agc004 agc 简体版
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AGC004

A - Divide a Cuboid

看哪一維是偶數，答案是0，不然是三個數兩兩組合相乘中最小的那個node

#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 1000005
//#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 A,B,C;
void Solve() {
    read(A);read(B);read(C);
    if(A % 2 == 0 || B % 2 == 0 || C % 2 == 0) {out(0);enter;}
    else {
    out(min(min(1LL * A * B,1LL * B * C),1LL * A * C));enter;
    }
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

B - Colorful Slimes

枚舉第二種操做的次數是k，每一個slime得到的時間能夠成爲前k個（循環）中最小的那個時間c++

#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);
}
int N,len[MAXN * 2];
int64 x,a[MAXN * 2],st[MAXN * 2][15];
int64 query(int l,int r) {
    int t = len[r - l + 1];
    return min(st[l][t],st[r - (1 << t) + 1][t]);
}
void Solve() {
    read(N);read(x);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);a[i + N] = a[i];st[i][0] = a[i];st[i + N][0] = a[i];}
    for(int j = 1 ; j <= 13 ; ++j) {
    for(int i = 1 ; i <= 2 * N ; ++i) {
        if(i + (1 << j) - 1 > 2 * N) break;
        st[i][j] = min(st[i][j - 1],st[i + (1 << j - 1)][j - 1]);
    }
    }
    for(int i = 2 ; i <= 2 * N ; ++i) len[i] = len[i / 2] + 1;
    int64 ans = 1e18;
    for(int i = 0 ; i <= N ; ++i) {
    int64 tmp = i * x;
    for(int j = 1 ; j <= N ; ++j) {
        tmp += query(j + N - i,j + N);
    }
    ans = min(ans,tmp);
    }
    out(ans);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

C - AND Grid

直接最上列藍色，最下列紅色，中間奇數列（除了頂端和底端）紅色，偶數列藍色，再把紫色的分配給兩個，這時候必定兩個都連通且只有紫色部分重合ide

#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 505
//#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;
char s[MAXN][MAXN];
char a[MAXN][MAXN],b[MAXN][MAXN];
void Solve() {
    read(H);read(W);
    for(int i = 1 ; i <= H ; ++i) scanf("%s",s[i] + 1);
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) {
        a[i][j] = b[i][j] = s[i][j];
    }
    }
    for(int i = 1 ; i <= W ; ++i) {a[1][i] = '#';b[H][i] = '#';}
    for(int j = 1 ; j <= W ; ++j) {
    if(j & 1) {
        for(int i = 1 ; i < H ; ++i) a[i][j] = '#';
    }
    else {
        for(int i = 2 ; i <= H ; ++i) b[i][j] = '#';
    }
    }
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) {
        putchar(a[i][j]);
    }
    enter;
    }
    enter;
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) {
        putchar(b[i][j]);
    }
    enter;
    }
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Teleporter

初始是一個基環內向樹，若是有一個環大於1不管如何存在一個點使得k步以後不在1，因而把1的邊連向本身，而後就變成了一棵樹，dp，每遇到大於k步的時候都把那個點的邊直接連到1便可函數

#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 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);
}
struct node {
    int to,next;
}E[MAXN * 2];
int N,K,head[MAXN],sumE;
int a[MAXN],ans;
int rem[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u) {
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    dfs(v);
    if(rem[v] == K - 1 && u != 1) {++ans;rem[v] = -1;}
    rem[u] = max(rem[u],rem[v] + 1);
    }
}
void Solve() {
    read(N);read(K);
    for(int i = 1 ; i <= N ; ++i) {
    read(a[i]);
    }
    if(a[1] != 1) ++ans;
    for(int i = 2 ; i <= N ; ++i) {
    add(a[i],i);
    }
    dfs(1);
    out(ans);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Salvage Robots

設\(dp[i][j][k][t]\)表示Exit向上最多延伸i，向下最多延伸k，向左最多延伸j，向右最多延伸tspa

此時下邊i行，上面k行，左邊t列，右邊j列，全都不合法，若是咱們要向上下左右擴展某一行某一列，那麼須要在合法的區域內選擇這一行或這一列的機器人個數code

#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 H,W;
char s[105][105];
int sum[105][105],a,b;
int dp[2][105][105][105],ans;
int query(int x1,int y1,int x2,int y2) {
    
    x2 = min(H,x2);y2 = min(W,y2);
    if(x1 > x2 || y1 > y2) return 0;
    return sum[x2][y2] - sum[x2][y1 - 1] - sum[x1 - 1][y2] + sum[x1 - 1][y1 - 1];
}
void update(int &x,int y) {
    x = max(x,y);
}
void Solve() {
    read(H);read(W);
    for(int i = 1 ; i <= H ; ++i) scanf("%s",s[i] + 1);
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) {
        sum[i][j] = (s[i][j] == 'o');
        sum[i][j] += sum[i][j - 1];
        if(s[i][j] == 'E') {a = i;b = j;}
    }
    }
    for(int i = 1 ; i <= H ; ++i) {
    for(int j = 1 ; j <= W ; ++j) {
        sum[i][j] += sum[i - 1][j];
    }
    }
    int cur = 0;
    dp[0][0][0][0] = 0;
    for(int i = 0 ; i < a ; ++i) {
    memset(dp[cur ^ 1],0,sizeof(dp[cur ^ 1]));
    for(int j = 0 ; j < b ; ++j) {
        for(int k = 0 ; k <= H - a ; ++k) {
        for(int t = 0 ; t <= W - b ; ++t) {
            ans = max(ans,dp[cur][j][k][t]);
            int tmp = dp[cur][j][k][t];
            if(k < a - i - 1) tmp += query(a - i - 1,max(b - j,t + 1),a - i - 1,min(b + t,W - j));
            update(dp[cur ^ 1][j][k][t],tmp);
            tmp = dp[cur][j][k][t];
            if(t < b - j - 1) tmp += query(max(a - i,k + 1),b - j - 1,min(a + k,H - i),b - j - 1);
            update(dp[cur][j + 1][k][t],tmp);
            tmp = dp[cur][j][k][t];
            if(H - i >= a + k + 1) tmp += query(a + k + 1,max(b - j,t + 1),a + k + 1,min(b + t,W - j));
            update(dp[cur][j][k + 1][t],tmp);
            tmp = dp[cur][j][k][t];
            if(W - j >= b + t + 1) tmp += query(max(a - i,k + 1),b + t + 1,min(a + k,H - i),b + t + 1);
            update(dp[cur][j][k][t + 1],tmp);
        }
        }
    }
    cur ^= 1;
    }
    out(ans);enter;
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - Namori

十分鐘內輕鬆水過1500（我吊打yutaka了），痛罵出題人竟然給這麼簡單的部分分1500而後E題1400？肝E註定失敗？ci

而後開始水基環樹，大膽猜結論以爲斷邊可行，可是樣例告訴我奇環不行，偶環過了不少可是沒全過（用assert判的奇環偶環）rem

實際上轉化一下問題，樹確定是個二分圖，咱們把它二分圖染色，咱們要作的就是把初始的染黑和染白交換位置，若是黑白個數不相等確定不行get

而後咱們只須要一個簡單的dp從下往上計算就能夠了it

對於偶環樹呢，咱們抽出環中的一條邊\((u\rightarrow v)\)而後認爲這個邊上走過的黑色標記有x個（x是負的那麼就是從v到u）

每條邊走過的黑色節點均可以用含x的代數式表示，答案是取某個x這些代數式絕對值和的最小值

是個凹函數，能夠三分

對於奇環樹呢，咱們至關於多了兩個操做，就是選出環上的一條邊做爲特殊邊，邊上相鄰兩點能夠刪掉兩個標記或者同時增長兩個標記

記錄下環上每一個點缺乏多少，而後從兩個端點分別讓這些標記開始走便可

#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 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);
}
struct node {
    int to,next;
}E[MAXN * 2];
int N,head[MAXN],sumE,M,fa[MAXN],dep[MAXN],st,ed;
int64 dp[MAXN][2],ans;
bool vis[MAXN];
int c[MAXN],tot,col[MAXN],rec[2],ned[MAXN];
pii val[MAXN];
vector<pii > f;
vector<int> dfn;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u) {
    dp[u][0] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa[u]) {
        fa[v] = u;
        dfs(v);
        ans += dp[v][0] + dp[v][1];
        dp[u][0] += dp[v][1];
        dp[u][1] += dp[v][0];
    }
    }
    int t = min(dp[u][0],dp[u][1]);
    dp[u][0] -= t;dp[u][1] -= t;
}
void find_circle(int u) {
    dep[u] = dep[fa[u]] + 1;
    vis[u] = 1;
    dfn.pb(u);
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(!vis[v]) {
        fa[v] = u;
        col[v] = col[u] ^ 1;
        find_circle(v);
    }
    else if(dep[v] < dep[u] && v != fa[u]) {
        
        st = u;ed = v;
    }
    }
}
int64 Calc(int64 x) {
    int64 res = 0;
    for(auto t : f) {
    res += abs(t.fi * x + t.se);
    }
    return res;
}
void EvenCircle() {
    for(int i = dfn.size() - 1 ; i >= 0 ; --i) {
    int u = dfn[i];
    pii p = val[u];
    if(u == st) {
        f.pb(mp(1,0));
        p.fi -= 1;
    }
    else if(u == ed) p.fi += 1;
    if(col[u] == 0) p.se--;
    else p.se++;
    if(fa[u]) {
        f.pb(p);
        val[fa[u]].fi += p.fi;val[fa[u]].se += p.se;
    }
    }
    int64 L = -N,R = N;
    while(R - L >= 3) {
    int64 Lb = L + (R - L) / 3;
    int64 Rb = R - (R - L) / 3;
    if(Calc(Lb) > Calc(Rb)) L = Lb;
    else R = Rb;
    }
    int64 res = 1e18;
    for(int64 i = L ; i <= R ; ++i) res = min(res,Calc(i));
    out(res);enter;
}
void Calc_need(int u) {
    if(col[u]) ned[u] = -1;
    else ned[u] = 1;
    vis[u] = 1;
    for(int i = head[u] ; i ;  i = E[i].next) {
    int v = E[i].to;
    if(!vis[v]) {
        Calc_need(v);
        ans += abs(ned[v]);
        ned[u] += ned[v];
    }
    }
}
void OddCircle() {
    int64 add = (rec[1] - rec[0]) / 2;
    memset(vis,0,sizeof(vis));
    for(int i = 1 ; i <= tot ; ++i) {
    vis[c[i]] = 1;
    }
    for(int i = 1 ; i <= tot ; ++i) {
    Calc_need(c[i]);
    }
    ned[c[1]] += add;
    ned[c[tot]] += add;
    for(int i = 2 ; i <= tot ; ++i) {
    ans += abs(ned[c[i - 1]]);
    ned[c[i]] += ned[c[i - 1]];
    ned[c[i - 1]] = 0;
    }
    for(int i = tot - 1 ; i >= 1 ; --i) {
    ans += abs(ned[c[i + 1]]);
    ned[c[i]] += ned[c[i + 1]];
    }
    out(ans + abs(add));enter;
}
void Process() {
    find_circle(1);
    for(int i = 1 ; i <= N ; ++i) {
    rec[col[i]]++;
    }
    int p = st;
    while(1) {
    c[++tot] = p;
    if(p == ed) break;
    p = fa[p];
    }
    if(tot % 2 == 0) {
    if(rec[0] != rec[1]) {puts("-1");return;}
    EvenCircle();
    }
    else {
    if((rec[0] ^ rec[1]) & 1) {puts("-1");return;}
    OddCircle();
    }
}
void Solve() {
    srand(time(0));
    read(N);read(M);
    int a,b;
    for(int i = 1 ; i <= M ; ++i) {
    read(a);read(b);
    add(a,b);add(b,a);
    }
    if(N & 1) {puts("-1");return;}
    if(M == N - 1) {
    dfs(1);
    if(dp[1][0] || dp[1][1]) {puts("-1");return;}
    out(ans);enter;
    }
    else Process();
}

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}
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