#8：二日合辑——6

BZOJ1001，裸网络流，对偶图做法比较有趣但在这道题上好像有点花哨？

#include <bits/stdc++.h>
#define ri readint()
#define wi(x) writeint(x)
#define gc getchar()
#define pc(x) putchar(x)
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define init(a, b) memset(a, b, sizeof(a))
using namespace std;

const int inf = 1e9;
const int maxn = 1e6 + 5;
const int maxm = 6e6 + 1e4;

struct Edge {
    int to, cost, nxt;
}e[maxm];
int head[maxn], d[maxn], tot = 1;
int n, m, st, ed;
queue<int> Q;

inline int readint() {
    int x = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)
        x = x * 10 + c - 48;
    return x * s;
}

inline void writeint(int x) {
    if (x > 9)    writeint(x/10);
    pc(x%10 + 48);
}

inline void add(int x, int y, int z) {
    e[++tot].to = y, e[tot].cost = z, e[tot].nxt = head[x], head[x] = tot;
    e[++tot].to = x, e[tot].cost = z, e[tot].nxt = head[y], head[y] = tot;
}

bool bfs() {
    init(d, 0);
    while (Q.size())    Q.pop();

    Q.push(st), d[st] = 1;
    while (Q.size()) {
        int x = Q.front(); Q.pop();
        for (int i = head[x]; i; i = e[i].nxt) {
            if (e[i].cost && !d[e[i].to]) {
                Q.push(e[i].to);
                d[e[i].to] = d[x] + 1;
                if (e[i].to == ed)    return true;
            }
        }
    }

    return false;
}

int dinic(int x, int flow) {
    if (x == ed)    return flow;
    int rest = flow;
    for (int i = head[x]; i && rest; i = e[i].nxt) {
        if (e[i].cost && d[e[i].to] == d[x] + 1) {
            int k = dinic(e[i].to, min(rest, e[i].cost));
            if (!k)    d[e[i].to] = 0;
            e[i].cost -= k;
            e[i ^ 1].cost += k;
            rest -= k;
        }
    }

    return flow - rest;
}

int solve() {
    int maxflow = 0;
    st = 1, ed = n * m;
    while (bfs())    maxflow += dinic(st, inf);
    return maxflow;
}

int main() {
    n = ri, m = ri;
    rep(i, 1, n)
        rep(j, 1, m-1) {
            int c = ri;
            int x = (i-1)*m+j, y = x+1;
            add(x, y, c);
        }
    rep(i, 1, n-1)
        rep(j, 1, m) {
            int c = ri;
            int x = (i-1)*m+j, y = i*m+j;
            add(x, y, c);
        }
    rep(i, 1, n-1)
        rep(j, 1, m-1) {
            int c = ri;
            int x = (i-1)*m+j, y = i*m+j+1;
            add(x, y, c);
        }

    wi(solve());
    return 0;
}

 

BZOJ1004，刻骨铭心的burnside。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

typedef long long ll;
int Sr, Sb, Sg, m, p;
int n, C[65], ans;
int f[25][25][25];
vector<int> Equal;
bool mark[65];

int cal() {
    memset(mark, false, sizeof(mark));
    memset(f, 0, sizeof(f));
    Equal.clear();

    for (int i = 1; i <= n; i++) {
        if (!mark[i]) {//如果还不属于某个循环节
            int k = 0;
            for (int j = i; !mark[j]; j = C[j]) {
                mark[j] = true;
                k++;
            }
            Equal.push_back(k);//这个循环节里共有多少个元素
        }
    }

    f[0][0][0] = 1;
    for (int i = 0; i < Equal.size(); i++)//枚举每个循环节
        for (int jr = Sr; ~jr; jr--)
            for (int jb = Sb; ~jb; jb--)
                for (int jg = Sg; ~jg; jg--) {
                    int size = Equal[i];
                    if (size <= jr)
                        f[jr][jb][jg] = (f[jr][jb][jg] + f[jr - size][jb][jg]) % p;
                    if (size <= jb)
                        f[jr][jb][jg] = (f[jr][jb][jg] + f[jr][jb - size][jg]) % p;
                    if (size <= jg)
                        f[jr][jb][jg] = (f[jr][jb][jg] + f[jr][jb][jg - size]) % p;
                }
    return f[Sr][Sb][Sg];
}

int ksm(int a, int b, int mod) {
    int res = 1;
    for (; b; b >>= 1) {
        if (b & 1)    res = (ll)res * a % mod;
        a = (ll)a * a % mod;
    }
    return res;
}

int main() {
    cin >> Sr >> Sb >> Sg >> m >> p;
    n = Sr + Sb + Sg;

    for (int i = 1; i <= m; i++) {//每个置换
        for (int i = 1; i <= n; i++)
            cin >> C[i];
        ans = (ans + cal()) % p;
    }
    for (int i = 1; i <= n; i++)//单位元置换
        C[i] = i;//e * e = e，即完全不动
    ans = (ans + cal()) % p;

    cout << ans * ksm(m+1, p-2, p) % p << endl;//(ans/(m+1))%p
    return 0;
}

 

UVa10294，polya的基本应用题。

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int n, t;
ll Pow[51], a, b;

int main() {
    while (~scanf("%d%d", &n, &t)) {
        Pow[0] = 1ll;
        for (int i = 1; i <= n; i++)
            Pow[i] = Pow[i-1] * t;

        a = 0ll;
        for (int i = 1; i <= n; i++)//转n个就是没转
            a += Pow[__gcd(i, n)];

        if (n & 1)    b = n * Pow[n/2 + 1];
        else    b = n / 2 * (Pow[n/2 + 1] + Pow[n/2]);
        
        printf("%lld %lld\n", a / n, (a+b) / (2*n));
    }
    return 0;
}

 

UVALive3641，有置换的结论，对于置换A²，循环节奇数长度相乘还是原长，偶数长度相乘分裂为原来的一半相乘。

#include <bits/stdc++.h>
using namespace std;

int T;
char S[27];
int cnt[27];
bool vis[26];

int main() {
    for (cin >> T; T; T--) {
        cin >> S;
        memset(vis, 0, sizeof(vis));
        memset(cnt, 0, sizeof(cnt));

        for (int i = 0; S[i]; i++) {
            if (!vis[i]) {
                int k = 0;
                for (int j = i; !vis[j]; j = S[j] - 'A') {
                    k++;
                    vis[j] = true;
                }
                cnt[k]++;
            }
        }

        bool flag = true;
        for (int i = 2; i <= 26; i += 2)
            if (cnt[i] & 1)
                flag = false;

        if (flag)    puts("Yes");
        else    puts("No");
    }
    return 0;
}

 

UVa11077，一是找到“交换即在同一置换循环节里”这一本质，二是递推。私以为蓝书递推解释错了，但代码是对的。f[i-1][j-1]代表了循环节个数不变还是i-j个但是这些循环节加起来共有i-1个位置可供i去放所以乘i-1；还有一种情况就是i-1-j个循环节然后i独立成节。

#include <bits/stdc++.h>
#define ull unsigned long long
using namespace std;

int n, k;
ull f[25][25];

int main() {
    for (int i = 1; i <= 21; i++) {
        f[i][0] = 1;
        for (int j = 1; j < i; j++) {
            f[i][j] = f[i-1][j] + f[i-1][j-1] * (i-1);
        }
    }

    while (~scanf("%d%d", &n, &k) && n) {
        printf("%llu\n", f[n][k]);
    }
    return 0;
}

 

UVALive3510，循环节lcm常见操作，but这波图像模拟真是emmmmm……

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1 << 10;
int n, n2, T;
int cur[maxn*maxn], ogri[maxn*maxn];
bool vis[maxn*maxn];

inline int id(int i, int j) {
    return i * n + j;
}

int Newid(int i, int j, const char *op) {
    if (op[0] == 'i')
        return id(i, j);
    else if (op[0] == 'r')
        return id(n-1-j, i);
    else if (op[0] == 's')
        return id(i, n-1-j);
    else if (op[0] == 'b' && op[1] == 'h')
        return i < n/2 ? id(i, j) : id(i, n-1-j);
    else if (op[0] == 'b' && op[1] == 'v')
        return i < n/2 ? id(i, j) : id(n/2+n-i-1, j);
    else if (op[0] == 'd')
        return i & 1 ? id(n/2+i/2, j) : id(i/2, j);
    else {
        int k = i/2;
        if (j < n/2)
            return i & 1 ? id(2*k, 2*j+1) : id(2*k, 2*j);
        else
            return i & 1 ? id(2*k+1, 2*(j-n/2)+1) : id(2*k+1, 2*(j-n/2));
    }    
}

void Apply(const char *op) {
    bool inv = op[strlen(op)-1] == '-';
    for (int i = 0; i < n*n; i++)    ogri[i] = cur[i];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) {
            int p = id(i, j), p2 = Newid(i, j, op);
            if (inv)    cur[p] = ogri[p2];
            else    cur[p2] = ogri[p];
        }
}

inline int __lcm(int a, int b) {
    return a / __gcd(a, b) * b;
}

int Solve(int m) {
    memset(vis, 0, sizeof(vis));
    int ans = 1;
    for (int i = 0; i < m; i++) {
        if (!vis[i]) {
            int len = 0;
            for (int j = i; !vis[j]; j = cur[j]) {
                vis[j] = true;
                len++;
            }
            ans = __lcm(ans, len);
        }
    }
    return ans;
}

int main() {
    for (scanf("%d%d", &T, &n); T--; n = n2) {
        for (int i = 0; i < n*n; i++)
            cur[i] = i;
        string s;
        vector<string> v;
        while (cin >> s) {
            if (isdigit(s[0])) {
                sscanf(s.c_str(), "%d", &n2);
                break;
            }
            v.push_back(s);
        }
        for (int i = v.size()-1; ~i; i--) {
            Apply(v[i].c_str());
        }

        printf("%d\n", Solve(n*n));
        if (T)    puts("");
    }
    return 0;
}