【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();
}
