【AtCoder】ARC093

C - Traveling Plan

相当于一个环，每次删掉i点到两边的距离，加上新相邻的两个点的距离

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
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) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int64 A[100005],ans;

int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(A[i]);
    for(int i = 1 ; i <= N + 1; ++i) {
	ans += abs(A[i] - A[i - 1]);
    }
    for(int i = 1 ; i <= N ; ++i) {
	int64 tmp = ans;
	tmp -= abs(A[i] - A[i - 1]) + abs(A[i] - A[i + 1]);
	tmp += abs(A[i - 1] - A[i + 1]);
	out(tmp);enter;
    }
}

D - Grid Components

每次这样

先拎出一个黑联通块和一个白联通块

黑黑黑黑黑黑

白黑白黑白黑

黑黑黑黑黑黑

白黑白黑白黑

黑黑黑黑黑黑

这样每两行50个往下消，白的构造黑的同理

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
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) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
int A,B,w = 100,h = 0;
char a[105][105];
void Calc(int R,char W,char B)  {
    ++h;
    for(int i = 1 ; i <= w ; ++i) a[h][i] = B;
    if(!R) return;
    while(R >= 50) {
	R -= 50;
	for(int i = 1 ; i <= w ; ++i) {
	    if(i & 1) a[h + 1][i] = W;
	    else a[h + 1][i] = B;
	}
	for(int i = 1 ; i <= w ; ++i) a[h + 2][i] = B;
	h += 2;
    }
    if(R) {
	int t = 1;
	while(R--) {
	    a[h + 1][t] = W;
	    a[h + 1][t + 1] = B;
	    t += 2;
	}
	for(int i = t ; i <= w ; ++i) a[h + 1][i] = B;
	for(int i = 1 ; i <= w ; ++i) a[h + 2][i] = B;
	h += 2;
    }
}
void Solve() {
    read(A);read(B);
    --A;--B;
    Calc(A,'.','#');Calc(B,'#','.');
    out(h);space;out(w);enter;
    for(int i = 1 ; i <= h ; ++i) {
	for(int j = 1 ; j <= w ; ++j) {
	    putchar(a[i][j]);
	}
	enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Bichrome Spanning Tree

我们随意求一个生成树\(T\)出来，然后给每个非树边求一个\(diff(e)\)表示边的值减去\(u,v\)上路径最大值

我们设\(D = X - T\)

\(equal\)是\(diff(e) == D\)的个数

\(upper\)是\(diff(e) > D\)的个数

\(lower\)是\(diff(e) < D\)的个数

然后对于\(D < 0\)无解

对于\(D = 0\)

我们可以对于树上的边两种颜色染色\((2^{N - 1} - 2)2^{M - N + 1}\)

如果树上的边都是一种颜色，那么答案是\(2(2^{equal} - 1)2^{upper}\)，就是\(diff(e) == D\)的边至少有一个不同颜色的，大于的边随意染色

对于\(D > 0\)

我们对于树上的边和\(lower\)边必须用同一种颜色染色

答案是\(2(2^{equal} - 1)2^{upper}\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define MAXN 2005
//#define ivorysi
using namespace std;
typedef long long int64;
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) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M,bl[MAXN],fa[MAXN],dep[MAXN],up,eq;
int64 X,T,faE[MAXN];
bool vis[MAXN];
struct Edge {
    int to,next;int64 val;
}E[MAXN * 2];
int head[MAXN],sumE;
struct node {
    int u,v;int64 val;
    friend bool operator < (const node &a,const node &b) {
	return a.val < b.val;
    }
}Ed[MAXN];
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;
}
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;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void add(int u,int v,int64 c) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].val = c;
    head[u] = sumE;
}
void dfs(int u) {
    dep[u] = dep[fa[u]] + 1;
    for(int i = head[u] ; i ; i = E[i].next) {
	int v = E[i].to;
	if(v != fa[u]) {
	    fa[v] = u;
	    faE[v] = E[i].val;
	    dfs(v);
	}
    }
}
int getfa(int u) {
    return bl[u] == u ? u : bl[u] = getfa(bl[u]);
}
int64 Query(int u,int v) {
    if(dep[u] < dep[v]) swap(u,v);
    int64 res = 0;
    while(dep[u] > dep[v]) {
	res = max(res,faE[u]);
	u = fa[u];
    }
    if(u == v) return res;
    while(u != v) {
	res = max(res,faE[u]);
	res = max(res,faE[v]);
	u = fa[u];v = fa[v];
    }
    return res;
}
void Solve() {
    read(N);read(M);read(X);
    for(int i = 1; i <= M ; ++i) {
	read(Ed[i].u);read(Ed[i].v);read(Ed[i].val);
    }
    for(int i = 1 ; i <= N ; ++i) bl[i] = i;
    sort(Ed + 1,Ed + M + 1);
    for(int i = 1 ; i <= M ; ++i) {
	if(getfa(Ed[i].u) != getfa(Ed[i].v)) {
	    vis[i] = 1;
	    T += Ed[i].val;
	    add(Ed[i].u,Ed[i].v,Ed[i].val);
	    add(Ed[i].v,Ed[i].u,Ed[i].val);
	    bl[getfa(Ed[i].u)] = getfa(Ed[i].v);
	}
    }
    dfs(1);
    for(int i = 1 ; i <= M ; ++i) {
	if(!vis[i]) {
	    int64 d = Ed[i].val - Query(Ed[i].v,Ed[i].u);
	    if(d == X - T) ++eq;
	    else if(d > X - T) ++up;
	}
    }
    if(X < T) {puts("0");return;}
    else if(X == T) {
	int ans = 0;
	update(ans,mul(inc(fpow(2,N - 1),MOD - 2),fpow(2,M - N + 1)));
	update(ans,mul(mul(2,inc(fpow(2,eq),MOD - 1)),fpow(2,up)));
	out(ans);enter;
    }
    else {
	int ans = 0;
	update(ans,mul(mul(2,inc(fpow(2,eq),MOD - 1)),fpow(2,up)));
	out(ans);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Dark Horse

我们把1当做放在第一个，之后的答案乘上\(2^N\)

答案显然就是\(2^0,2^1,2^2...2^(N - 1)\)大小的集合的最小值不为给定的\(M\)个数之一

我们计算\(f(S)\)表示\(S\)所代表的集合的最小值都是\(M\)个数之一，剩下的随意的方案数

答案就是容斥\(\sum(-1)^{|S|}f(S)\)

设\(dp[i][S]\)表示考虑到第\(i\)大的\(A\)，然后集合为\(S\)的都填满且最小值为\(M\)个数之一的方案数

因为从大到小填数可以很容易算出来当前集合还有几个可以用的

\(f[S] = dp[M][S]\)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
//#define ivorysi
using namespace std;
typedef long long int64;
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) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int fac[(1 << 16) + 5],invfac[(1 << 16) + 5],inv[(1 << 16) + 5],N,M;
int A[25],f[(1 << 16) + 5],dp[17][(1 << 16) + 5],cnt[(1 << 16) + 5];
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;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
int lowbit(int x) {
    return x & (-x);
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= M ; ++i) read(A[i]);
    inv[1] = 1;
    for(int i = 2 ; i <= (1 << N) ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    fac[0] = invfac[0] = 1;
    for(int i = 1 ; i <= (1 << N) ; ++i) {
	fac[i] = mul(fac[i - 1],i);
	invfac[i] = mul(invfac[i - 1],inv[i]);
    }
    sort(A + 1,A + M + 1);
    dp[0][0] = 1;
    for(int i = 1 ; i <= M ; ++i) {
	int t = M - i + 1;
	for(int S = 0 ; S < (1 << N) ; ++S) {
	    for(int j = 0 ; j < N ; ++j) {
		if(!(S >> j & 1)) {
		    update(dp[i][S ^ (1 << j)],mul(dp[i - 1][S],mul(C((1 << N) - A[t] - S,(1 << j) - 1),fac[1 << j])));
		}
	    }
	    update(dp[i][S],dp[i - 1][S]);
	}
    }
    int ans = 0;
    for(int S = 0 ; S < (1 << N) ; ++S) {
	if(S) cnt[S] = cnt[S - lowbit(S)] + 1;
	int t = mul(dp[M][S],fac[(1 << N) - 1 - S]);
	if(cnt[S] & 1) update(ans,MOD - t);
	else update(ans,t);
    }
    ans = mul(ans,1 << N);
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}