HDU 2254 奥运（矩阵快速幂+二分等比序列求和）

ACM

题目地址：HDU 2254 奥运

题意：

中问题不解释。

分析：

根据floyd的算法，矩阵的k次方表示这个矩阵走了k步。

所以k天后就算矩阵的k次方。

这样就变成：初始矩阵的^[t1,t2]这个区间内的v[v1][v2]的和。

所以就是二分等比序列求和上场的时候了。

跟HDU 1588 Gauss Fibonacci的算法一样。

代码：

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  Blog:        http://blog.csdn.net/hcbbt
*  File:        2254.cpp
*  Create Date: 2014-08-04 10:52:29
*  Descripton:  matrix, floyd
*/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
typedef long long ll;

const int N = 20;
const int SIZE = 32;        // max size of the matrix
const int MOD = 2008;

int n, k, p1, p2;
ll v1, v2, t1, t2;
map<int, int> mp;

struct Mat{
    int n;
    ll v[SIZE][SIZE];    // value of matrix

    Mat(int _n = SIZE) {
        n = _n;
    }

    void init(ll _v = 0) {
        memset(v, 0, sizeof(v));
        if (_v)
            repf (i, 0, n - 1)
                v[i][i] = _v;
    }

    void output() {
        repf (i, 0, n - 1) {
            repf (j, 0, n - 1)
                printf("%lld ", v[i][j]);
            puts("");
        }
        puts("");
    }
} a, b;

Mat operator * (Mat a, Mat b) {
    Mat c(a.n);
    repf (i, 0, a.n - 1) {
        repf (j, 0, a.n - 1) {
            c.v[i][j] = 0;
            repf (k, 0, a.n - 1) {
                c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;
                c.v[i][j] %= MOD;
            }
        }
    }
    return c;
}

Mat operator ^ (Mat a, ll k) {
    Mat c(a.n);
    c.init(1);
    while (k) {
        if (k&1) c = a * c;
        a = a * a;
        k >>= 1;
    }
    return c;
}

Mat operator + (Mat a, Mat b) {
    Mat c(a.n);
    repf (i, 0, a.n - 1)
        repf (j, 0, a.n - 1)
            c.v[i][j] = (b.v[i][j] + a.v[i][j]) % MOD;
    return c;
}

Mat operator + (Mat a, ll b) {
    Mat c = a;
    repf (i, 0, a.n - 1)
        c.v[i][i] = (a.v[i][i] + b) % MOD;
    return c;
}

Mat calc(Mat a, int n) {
    if (n == 1)
        return a;
    if (n&1)
        return (a^n) + calc(a, n - 1);
    else
        return calc(a, n/2) * ((a^(n/2)) + 1);
}

int main() {
    while (~scanf("%d", &n)) {
        a.init();
        mp.clear();
        int cnt = 0;
        while (n--) {
            scanf("%d%d", &p1, &p2);
            if (mp.find(p1) == mp.end())
                p1 = mp[p1] = cnt++;
            else
                p1 = mp[p1];
            if (mp.find(p2) == mp.end())
                p2 = mp[p2] = cnt++;
            else
                p2 = mp[p2];
            a.v[p1][p2]++;
        }
        a.n = cnt;

        scanf("%d", &k);
        while (k--) {
            scanf("%lld%lld%lld%lld", &v1, &v2, &t1, &t2);
            if (mp.find(v1) == mp.end() || mp.find(v2) == mp.end()) {
                puts("0");
                continue;
            }
            v1 = mp[v1];
            v2 = mp[v2];
            if (t1 > t2)
                swap(t1, t2);
            if (t1 == 0) {
                if (t2 == 0)
                    puts("0");
                else
                    printf("%lld\n", calc(a, t2).v[v1][v2]);
            }
            else if (t1 == 1)
                printf("%lld\n", calc(a, t2).v[v1][v2]);
            else {
                printf("%lld\n", ((calc(a, t2).v[v1][v2] - calc(a, t1 - 1).v[v1][v2]) + MOD) % MOD);
            }
        }
    }
    return 0;
}

HDU 2254 奥运（矩阵快速幂+二分等比序列求和）