C++博客- -随笔分类-Algorithm

SGU282 Isomorphism-polya计数

Sun, 11 Oct 2009 07:12:00 GMT

题目：http://acm.sgu.ru/problem.php?contest=0&problem=282
大意：对N阶完全图的边染色，求在置换群的作用下，不等价的染色方法数，共有m中颜色， N <= 54，结果模上质数p
思路：每个置换分解为若干循环结，长度为i的循环结有ni个,则sigma{i * ni} = N. 每一个置换唯一对应一个n元组(n1, n2, ... , nN), 每个n元组可由N!/pi{i^ni}/pi{ni!}个置换得到， (pi{}, sigma{}, 分别表示求积和求和)。分别计算循环结内部子完全图中等价边类的个数([L/2])，和循环结之间的等价边类的个数(gcd(L1, L2)), 求出总等价边类个数sum，得出每个n元组下对应的染色m^sum。枚举所有n元组。
细节：枚举所有n元组的代价很高，需要对一些量预处理，作常数优化。

1 #include <iostream>
2
3 using namespace std;
4 typedef long long int64;
5 const int maxn = 64;
6 int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
7 int n, m, p, x[maxn], d[maxn][maxn], cn2[maxn];//x为n元组，d[i][j] = gcd(i, j), cn2[i] = i * (i - 1) / 2
8 int64 rv[maxn], fac[maxn], facv[maxn];//rv[i] = inv(i, p), fac[i] = i!, facv = inv(i!, p)
9 int64 ans = 0;
10 int64 pw(int64 a, int64 b, int64 M) {
11     if(b == 0) return 1;
12     int64 tmp = pw(a, b >> 1, M);
13     tmp = tmp * tmp % M;
14     if(b & 1) return (tmp * a) % M;
15     return tmp;
16 }
17 int64 inv(int64 a, int64 b) { return pw(a, b - 2, b); }//a模质数b的逆元
18 void init() {
19     rv[1] = fac[0] = 1;
20     for(int i = 2; i <= n; ++ i) rv[i] = inv(i, p);
21     for(int i = 1; i <= n; ++ i) fac[i] = i * fac[i - 1] % p;
22     for(int i = 0; i <= n; ++ i) facv[i] = inv(fac[i], p);
23 }
24 int64 cal() {
25     int64 sum = 0, toadd = 1;
26     for(int i = 1; i <= n; ++ i) sum += i / 2 * x[i] + cn2[x[i]] * i;
27     for(int i = 1; i <= n; ++ i) for(int j = i + 1; j <= n; ++ j) if(x[i] && x[j])
28         sum += x[i] * x[j] * d[i][j];
29     for(int i = 1; i <= n; ++ i) {
30         int64 tmp = rv[i];
31         for(int j = 0; j < x[i]; ++ j)
32             toadd = toadd * tmp % p;
33         toadd = toadd * facv[x[i]] % p;
34     }
35     toadd = toadd * pw(m, sum, p) % p;
36     return toadd;
37 }
38 void dfs(int t, int sum) {//枚举所有n元组
39     if(sum == n) {
40         ans = (ans + cal()) % p;
41         return;
42     }
43     if(t > n) return;
44     for(int i = 0; i * t + sum <= n; ++ i) {
45         x[t] = i;
46         dfs(t + 1, t * i + sum);
47     }
48 }
49 int main()
50 {
51 //    freopen("in", "r", stdin);
52     for(int i = 1; i < maxn; ++ i)
53         for(int j = 1; j < maxn; ++ j)
54             d[i][j] = gcd(i, j);
55     for(int i = 0; i < maxn; ++ i) cn2[i] = (i - 1) * i / 2;
56     while(scanf("%d%d%d", &n, &m, &p) == 3) {
57         init();
58         ans = 0;
59         dfs(1, 0);
60         printf("%lld\n", ans);
61     }
62     return 0;
63 }
64

x 2009-10-11 15:12 发表评论

usaco-fence rail

Mon, 31 Aug 2009 12:30:00 GMT

题目链接：usaco-fence rails
题目大意：给定一些board的长度，一些rail的长度。求出最多能用board切出的rail数目。
数据范围: board数目<=50, rail数目<=1023, rail长度范围<=128
思路：二分+dfsid +剪枝

#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
const int maxr = 1024;
const int maxb = 64;

int rail[maxr], board[maxb], remain[maxb];
int nrail, nboard;
int sumr[maxr], sumb[maxb];
int choice[maxr];

bool cmp(int a, int b) {
    return a > b;
}

bool dfsid(int t, int n) {
    if(t == n) return true;
    int waste = 0;
    for(int i = 0; i < nboard; ++ i) if(remain[i] < rail[0]) waste += remain[i];
    if(waste + sumr[n - 1] > sumb[nboard - 1]) return false;//如果浪费太多剪掉
    int start = 0;
    if(t > 0 && rail[n - t - 1] == rail[n - t]) start = choice[t - 1];//如果当前rail和之前的rail长度相同直接从前一个rail尝试的board开始,注意虽然从小到大贪，但确定贪的个数之后却从大到小搜索
    for(int i = start; i < nboard; ++ i) {
        if(remain[i] >= rail[n - t - 1]) {
            remain[i] -= rail[n - t - 1];
            choice[t] = i;
            bool ret = dfsid(t + 1, n);
            remain[i] += rail[n - t - 1];
            choice[t] = -1;
            if(ret) return true;
        }
    }
    return false;
}
int main()
{
    freopen("fence8.in", "r", stdin);
    freopen("fence8.out", "w", stdout);
    scanf("%d", &nboard);
    for(int i = 0; i < nboard; ++ i) {
        scanf("%d", &board[i]);
        remain[i] = board[i];
    }
    sort(board, board + nboard, cmp);//board从大到小尝试
    scanf("%d", &nrail);
    for(int i = 0; i < nrail; ++ i) scanf("%d", &rail[i]);
    sort(rail, rail + nrail);//贪心:尽量取小的
    sumb[0] = board[0], sumr[0] = rail[0];
    for(int i = 1; i < nrail; ++ i) sumr[i] = sumr[i - 1] + rail[i];
    for(int i = 1; i < nboard; ++ i) sumb[i] = sumb[i - 1] + board[i];

    memset(choice, -1, sizeof(choice));
    int low = 0, high = nrail, ans = -1;

    while(sumr[high -  1] > sumb[nboard - 1]) high--;
    while(low <= high) {
        int mid = (low + high) >> 1;
        if(dfsid(0, mid)) {
            ans = mid;
            low = mid + 1;
        }
        else high = mid - 1;
    }
    printf("%d\n", ans);
    return 0;
}

x 2009-08-31 20:30 发表评论

pku-DNA repair trie图+dp

Fri, 14 Aug 2009 10:00:00 GMT

利用AC自动机的fail指针将所有的转移结果都存起来就成了trie图了。

#include <stdio.h>
#include <string.h>
#include <queue>
#include <iostream>
using namespace std;

const int maxnodes = 480;
const int maxlen = 1024;
const int maxchar = 4;

int child[maxnodes][maxchar];
bool danger[maxnodes];
int fail[maxnodes], nodes;
int dp[maxlen][maxnodes];
char str[maxlen];

inline int getid(char ch) {
    switch(ch) {
        case 'A':return 0;
        case 'G':return 1;
        case 'C':return 2;
        case 'T':return 3;
    }
    return -1;
}

void insert(char *words) {
    int p = 1;
    for(int i = 0; words[i]; ++ i) {
        int idx = getid(words[i]);
        if(child[p][idx] == 0)
            child[p][idx] = ++nodes;
        p = child[p][idx];
        if(danger[p]) break;
    }
    danger[p] = true;
}

bool build_trie() {

    int n;
    char words[30];
    scanf("%d", &n);
    if(n == 0) return false;

    nodes = 1;
    memset(child, 0, sizeof(child));
    memset(danger, 0, sizeof(danger));
    memset(fail, 0, sizeof(fail));

    for(int i = 0; i < n; ++ i) {
        scanf("%s", words);
        insert(words);
    }
    return true;
}

void build_graph() {
    queue<int> Q;
    for(int i = 0; i < maxchar; ++ i) {
        if(child[1][i] == 0) child[1][i] = 1;
        else {
            Q.push(child[1][i]);
            fail[child[1][i]] = 1;
        }
    }
    while(!Q.empty()) {
        int tmp = Q.front(); Q.pop();
        danger[tmp] = danger[tmp] || danger[fail[tmp]];
        if(!danger[tmp]) {
            for(int i = 0; i < maxchar; ++ i) {
                if(child[tmp][i] == 0) child[tmp][i] = child[fail[tmp]][i];
                else {
                    Q.push(child[tmp][i]);
                    fail[child[tmp][i]] = child[fail[tmp]][i];
                }
            }
        }
    }
}

void check(int &a, int b) { if(a == -1 || a > b) a = b; }

int main()
{
    int cases = 0;
    while(build_trie()) {
        printf("Case %d: ", ++cases);
        build_graph();
        scanf("%s", str);

        memset(dp, -1, sizeof(dp));
        if(!danger[child[1][getid(str[0])]]) dp[0][child[1][getid(str[0])]] = 0;
        for(int i = 0; i < maxchar; ++ i)
            if(getid(str[0]) != i && !danger[child[1][i]])
                dp[0][child[1][i]] = 1;
        for(int i = 1; str[i]; ++ i)
            for(int j = 1; j <= nodes; ++ j) {
                if(dp[i-1][j] != -1) {
                    int idx = getid(str[i]);
                    if(!danger[child[j][idx]])
                        check(dp[i][child[j][idx]], dp[i-1][j]);
                    for(int k = 0; k < maxchar; ++ k)
                        if(idx != k && !danger[child[j][k]])
                            check(dp[i][child[j][k]], dp[i-1][j] + 1);
                }
            }
        int ans = 1 << 20;
        int len = strlen(str);
        for(int i = 1; i <= nodes; ++ i)
            if(!danger[i] && dp[len-1][i] != -1) check(ans, dp[len-1][i]);
        if(ans == 1 << 20) ans  = -1;
        printf("%d\n",ans);
    }
    return 0;
}

x 2009-08-14 18:00 发表评论

hdu-addHP dp + 二分

Thu, 13 Aug 2009 11:21:00 GMT

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2891

#include <iostream>

using namespace std;
const int maxn = 10240;
int hp, n, m;
int maxhp[maxn], suma[maxn], attack[maxn], Tm[maxn], add[maxn];
int mint[maxn];

int main()
{
    while(scanf("%d%d%d", &n, &m, &hp) == 3) {
        memset(suma, 0, sizeof(suma));
        for(int i = 1; i <= n; ++ i) {
            scanf("%d", &attack[i]);
            suma[i] = suma[i - 1] + attack[i];
        }
        memset(mint, -1, sizeof(mint));
        for(int i = 0; i < m; ++ i) {
            scanf("%d%d", &Tm[i], &add[i]);
            if(mint[add[i]] == -1 || mint[add[i]] > Tm[i]) mint[add[i]] = Tm[i];
        }
        int top = 0;
        for(int i = 0; i < m; ++ i)
            if(mint[add[i]] == Tm[i]) {
                Tm[top] = Tm[i];
                add[top++] = add[i];
            }
        m = top;
        int low = 0, high = hp, res = 0;
        while(low <= high) {
            int mid = (low + high) >> 1;
//            cout << mid << endl << endl;
            memset(maxhp, 0, sizeof(maxhp));
            maxhp[0] = hp;
            bool ok = true;
            for(int i = 1; i <= n && ok; ++ i) {
                for(int j = 0; j < m && ok; ++ j) {
                    if(i >= Tm[j] && maxhp[i - Tm[j]] - suma[i] + suma[i - Tm[j]] >= mid) {
                        maxhp[i] >?= maxhp[i - Tm[j]] - suma[i] + suma[i - Tm[j]] + add[j];
                        maxhp[i]  hp;
                    }
                    else if(i < Tm[j] && maxhp[0] - suma[i] >= mid) maxhp[i] >?= hp - suma[i];
                }
                if(maxhp[i] <= mid) ok = false;
//                cout << maxhp[i] << ' ';
            }
//            cout << endl;
            if(ok) {
                res = mid;
                low = mid + 1;
            }
            else high = mid - 1;
        }
        if(res == 0) puts("Die");
        else printf("%d\n", res);
    }
    return 0;
}

x 2009-08-13 19:21 发表评论

mobius function

Wed, 01 Jul 2009 10:34:00 GMT

定义：sigma{ [d|m] * mu(d) } = [m = 1]
      m 1 2 3 4 5 6 7 8 9 10 11 12
mu(m) 1 -1 -1 0 -1 1 -1 0  0    1 -1    0
mobius 反演公式1: g(m) = sigma{[d | m] * f(d) } <=> f(m) = sigma{ [d | m]*mu(d)*g(m / d)}
            反演公式2：g(x) = sigma{[d >= 1] * f(x / d) } <=> f(x) = sigma{ [d >= 1] * g(x / d) }
利用定义的积性可知mu()积性：
            mu(m ) = (-1)^r m = p1*p2....pr;
                           0         某p^2 | m

类似的反演公式还有二项式反演。好像还有反演公式系统。

x 2009-07-01 18:34 发表评论

pibonacci

Tue, 28 Apr 2009 14:09:00 GMT

import java.io.*;
import java.util.*;
import java.math.*;
public class Main {

    static BigInteger cal(int n) {
        BigInteger res = new BigInteger("0");
        if(n < 4) return BigInteger.ONE;
        int[] r = new int[n];
        BigInteger[] A = new BigInteger[n];

        for(int i = 0; i <= n - 3; ++ i)
            r[i] = (int)((n-i-4)/Math.PI) + 1;

        A[0] = BigInteger.ONE;
        for(int i = 1; i <= n - 3; ++ i) {
            int m1 = i - 1 + r[i-1];
            int m2 = i + r[i];
            if(n - (i - 1) - r[i - 1] * Math.PI < 3) -- m1;
            if(n - i - r[i] * Math.PI < 3) -- m2;
            int d = m2 - m1;

            if(d == -1) A[i] = A[i-1].multiply(new BigInteger((m1 + 1 - i) * (m1 - i) + "")).divide(new BigInteger(i * m1 + ""));
            if(d == 0) A[i] = A[i-1].multiply(new BigInteger(m1 + 1 - i + "")).divide(new BigInteger(i + ""));
            if(d == 1) A[i] = A[i-1].multiply(new BigInteger(m1 + 1 + "")).divide(new BigInteger(i + ""));
            if(d == 2) A[i] = A[i-1].multiply(new BigInteger((m1 + 2) * (m1 + 1) + "")).divide(new BigInteger(i * (m1 + 2 - i) + ""));
        }

        for(int i = 0; i <= n - 3; ++ i)
             res = res.add(A[i]);
        return res;
    }
    static void myPrint(String s) {
        int len = s.length();
        for(int i = 0; i < len/50; ++ i) {
            for(int j = 0; j < 50; ++ j)
                System.out.print(s.charAt(i * 50 + j));
            System.out.println();
        }
        int ptr = len / 50 * 50;
        if(ptr < len) {
            for(int i = ptr; i < len; ++ i)
                System.out.print(s.charAt(i));
            System.out.println();
        }
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n;
        while(in.hasNext()){
            n = in.nextInt();
            if(n == -1) return;
            BigInteger res = cal(n);
            myPrint(res.toString());
        }
    }
}

x 2009-04-28 22:09 发表评论