【AtCoder】AGC004

AGC004

A - Divide a Cuboid

看哪一维是偶数，答案是0，否则是三个数两两组合相乘中最小的那个

#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

枚举第二种操作的次数是k，每个slime获得的时间可以成为前k个（循环）中最小的那个时间

#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

直接最上列蓝色，最下列红色，中间奇数列（除了顶端和底端）红色，偶数列蓝色，再把紫色的分配给两个，这时候一定两个都连通且只有紫色部分重合

#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

初始是一个基环内向树，如果有一个环大于1无论如何存在一个点使得k步之后不在1，于是把1的边连向自己，然后就变成了一棵树，dp，每遇到大于k步的时候都把那个点的边直接连到1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 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

设\(dp[i][j][k][t]\)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t

此时下边i行，上面k行，左边t列，右边j列，全都不合法，如果我们要向上下左右扩展某一行某一列，那么需要在合法的区域内选择这一行或这一列的机器人个数

#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

十分钟内轻松水过1500（我吊打yutaka了），痛骂出题人居然给这么简单的部分分1500然后E题1400？肝E注定失败？

然后开始水基环树，大胆猜结论觉得断边可行，但是样例告诉我奇环不行，偶环过了很多但是没全过（用assert判的奇环偶环）

实际上转化一下问题，树肯定是个二分图，我们把它二分图染色，我们要做的就是把初始的染黑和染白交换位置，如果黑白个数不相等肯定不行

然后我们只需要一个简单的dp从下往上计算就可以了

对于偶环树呢，我们抽出环中的一条边\((u\rightarrow v)\)然后认为这个边上走过的黑色标记有x个（x是负的那么就是从v到u）

每条边走过的黑色节点都可以用含x的代数式表示，答案是取某个x这些代数式绝对值和的最小值

是个凹函数，可以三分

对于奇环树呢，我们相当于多了两个操作，就是选出环上的一条边作为特殊边，边上相邻两点可以删掉两个标记或者同时增加两个标记

记录下环上每个点缺少多少，然后从两个端点分别让这些标记开始走即可

#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 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;
}