HDU 2254 奥运（矩阵高速幂+二分等比序列求和）

HDU 2254 奥运（矩阵高速幂+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：

依据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;
}