HDU 2254 奥运 Matrix Power


HDU 2254 奥运 Matrix Power
Problem Description
北京迎来了第一个奥运会,我们的欢呼声响彻中国大地,所以今年的奥运金牌 day day up!
比尔盖兹坐上鸟巢里,手里摇着小纸扇,看的不亦乐乎,被俺们健儿的顽强拼搏的精神深深的感动了。反正我的钱也多的没地方放了,他对自己说,我自己也来举办一个奥运会,看谁的更火。不过他的奥运会很特别:
1 参加人员必须是中国人;
2 至少会加法运算(因为要计算本人获得的金牌数)
他知道中国有很多的名胜古迹,他知道自己在t1 到 t2天内不可能把所有的地方都玩遍,所以他决定指定两个地方v1,v2,如果参赛员能计算出在t1到t2天(包括t1,t2)内从v1到v2共有多少种走法(每条道路走需要花一天的时间,且不能在某个城市停留,且t1=0时的走法数为0),那么他就会获得相应数量的金牌,城市的总数<=30,两个城市间可以有多条道路
,每条都视为是不同的。
Input
本题多个case,每个case:
输入一个数字n表示有n条道路 0<n<10000
接下来n行每行读入两个数字 p1,p2 表示城市p1到p2有道路,并不表示p2到p1有道路 (0<=p1,p2<2^32)
输入一个数字k表示有k个参赛人员
接下来k行,每行读入四个数据v1,v2,t1,t2 (0<=t1,t2<10000) 
Output
对于每组数据中的每个参赛人员输出一个整数表示他获得的金牌数(mod 2008)
Sample Input
6 1 2 1 3 2 3 3 2 3 1 2 1 3 1 2 0 0 1 2 1 100 4 8 3 50
Sample Output
0 1506 0
如果 a 到 b 有路,则map[a][b] = 1,否则map[a][b] = 0,譬如如果1→3,1→4,2→1,2→3,3→4,4→2,4→3有路,则可以构造矩阵:         
这样构造好矩阵之后,如果求从 1 到 2 走 5 天(每条路走一天),有几条路,就可以这样计算:                        
结果中,第 n 行第 m 列就是 n 到 m 要走 5 天的路数。
一张有向图的邻接矩阵A,A表示所有点之间路径长度为一的路径数量,A^n则表示路径长度为n的路径数量,故需要求某两点在(A^t1)~(A^t2)的路径数量之和

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。
HDU 1588 Gauss Fibonacci的算法一样。
struct Matrix {
    int  mat[N][N];
    int n, m; 
    Matrix() {//初始化
        n = m = N;
        memset(mat, 0, sizeof(mat));
    }     
    inline void init(int row, int column) {//初始矩阵大小
        n = row, m = column;
    }
    void init_e() {//单位矩阵
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                mat[i][j] = (i == j);
            }
        }
    }
}; 
//加法
Matrix operator + (Matrix a, Matrix b) {
    Matrix ret;
    ret.init(a.n, a.m);
    for (int i = 0; i < a.n; i++) {
        for (int j = 0; j < a.m; j++) {
            ret.mat[i][j] = a.mat[i][j] + b.mat[i][j];
            if (ret.mat[i][j] >= mod) {
                ret.mat[i][j] %= mod;
            }
        }
    }
    return ret;
} 
//n行m列  *  m行p列   =  n行p列
Matrix operator * (Matrix a, Matrix b) {
    Matrix ret;
    ret.init(a.n, b.m);
    for (int i = 0; i < a.n; i++) {
        for (int k = 0; k < a.m; k++) if (a.mat[i][k]) {
            for (int j = 0; j < b.m; j++) if (b.mat[k][j]) {
                ret.mat[i][j] = ret.mat[i][j] + a.mat[i][k] * b.mat[k][j]; 
                if (ret.mat[i][j] >= mod) {
                    ret.mat[i][j] %= mod;
                }//if
            }//for(j)
        }//for(k)
    }//for(i)
    return ret;
}
//求幂一般都是正方形矩阵,所以ret = a;
Matrix operator ^ (Matrix a, int b) {
    Matrix ret = a,  tmp = a;
    ret.init_e();
    for ( ; b; b >>= 1) {
        if (b & 1) {
            ret = ret * tmp;
        }
        tmp = tmp * tmp;
    }
    return ret;
}
//递归幂求和
//用二分法求矩阵和,递归实现 
Matrix Power_Sum1(Matrix a, int b) {
    Matrix ret = a;
    ret.init_e();
    if (b == 1) {
        return a;
    } else if (b & 1) {
        return (a ^ b) + Power_Sum1(a, b - 1);
    } else {
        return Power_Sum1(a, b >> 1) * ((a ^ (b >> 1)) + ret);
    }
}
//非递归幂求和
Matrix Power_Sum2(Matrix a, int b) {
    int k = 0 ,ss[32];
    Matrix tp1, tp2, ret;
    tp1 = tp2 = ret = a;
    ret.init_e();
    while (b) {
        ss[k++] = b & 1;
        b >>= 1;
    }
    for (int i = k - 2; i >= 0; i--) {
        tp1 = tp1 * (tp2 + ret);
        tp2 = tp2 * tp2;
        if (ss[i]) {
            tp2 = tp2 * a;
            tp1 = tp1 + tp2;
        }
    }
    return tp1;
}
int main(void) {
   int cas, n;
   int v1, v2, t1, t2, x, y;
   while (~scanf("%d", &cas)) {
       map <int, int> mp;//最多只有30条路,所以用map映射
       mp.clear();
 
       Matrix a, b, c;
 
       int cnt = 1;
       for (int i = 0; i < cas; i++) {
           scanf("%d %d", &x, &y);
           if (!mp[x]) {
               mp[x] = cnt++;
           }
           if (!mp[y]) {
               mp[y] = cnt++;
           }
           a.mat[mp[x]][mp[y]] ++;
       }
        
       a.init(cnt, cnt);
      // a.print();
       b.init(cnt, cnt);
       c.init(cnt, cnt);
 
       scanf("%d", &n);
       while (n--) {
           scanf("%d%d%d%d", &v1, &v2, &t1, &t2);    
           x = mp[v1];
           y = mp[v2];
           if (x == 0 || y == 0 || (t1 == 0 && t2 == 0)) {
               puts("0");
               continue;
           }
           int sum;
 
           if (t1 <= 1) {
               b = Power_Sum2(a, t2);
              // b.print();
               sum = b.mat[x][y] % mod;
               printf("%d\n",sum);
               continue;
           }
            b = Power_Sum2(a, t2);
            c = Power_Sum2(a, t1 - 1);
            sum = (b.mat[x][y] % mod - c.mat[x][y] % mod + mod) % mod;
            printf("%d\n", sum);
       }//while(n)
   }//while(cas)
   return 0;
}
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