【AtCoder】ARC093

C - Traveling Plan

至關於一個環，每次刪掉i點到兩邊的距離，加上新相鄰的兩個點的距離

代碼

#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

每次這樣
先拎出一個黑聯通塊和一個白聯通塊
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
這樣每兩行50個往下消，白的構造黑的同理

#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

咱們隨意求一個生成樹\(T\)出來，而後給每一個非樹邊求一個\(diff(e)\)表示邊的值減去\(u,v\)上路徑最大值
咱們設\(D = X - T\)
\(equal\)是\(diff(e) == D\)的個數
\(upper\)是\(diff(e) > D\)的個數
\(lower\)是\(diff(e) < D\)的個數

而後對於\(D < 0\)無解
對於\(D = 0\)
咱們能夠對於樹上的邊兩種顏色染色\((2^{N - 1} - 2)2^{M - N + 1}\)
若是樹上的邊都是一種顏色，那麼答案是\(2(2^{equal} - 1)2^{upper}\)，就是\(diff(e) == D\)的邊至少有一個不一樣顏色的，大於的邊隨意染色

對於\(D > 0\)
咱們對於樹上的邊和\(lower\)邊必須用同一種顏色染色
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

咱們把1當作放在第一個，以後的答案乘上\(2^N\)

答案顯然就是\(2^0,2^1,2^2...2^(N - 1)\)大小的集合的最小值不爲給定的\(M\)個數之一

咱們計算\(f(S)\)表示\(S\)所表明的集合的最小值都是\(M\)個數之一，剩下的隨意的方案數
答案就是容斥\(\sum(-1)^{|S|}f(S)\)

設\(dp[i][S]\)表示考慮到第\(i\)大的\(A\)，而後集合爲\(S\)的都填滿且最小值爲\(M\)個數之一的方案數
由於從大到小填數能夠很容易算出來當前集合還有幾個能夠用的
\(f[S] = dp[M][S]\)

#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();
}