HDU 2254

http://acm.hdu.edu.cn/showproblem.php?pid=2254

矩阵乘法两个经典问题的综合题，还要离散化和处理边界，好题啊好题

题意容易理解错，每一天是独立的，所以根据加法原理方案数是G^1+G^2+...+G^t

/*
此题要求 (G^1+G^2+...+G^t2)-(G^1+G^2+...+G^(t1-1)) 
求和的方法是再次二分,k=6时 
G + G^2 + G^3 + G^4 + G^5 + G^6 = G + G^2 + G^3 + G^3 * (G + G^2 + G^3) = (1 + G^3) * (G + G^2 + G^3) 
这样计算可以使k的规模减少一半，快速幂求出G^3后可递归计算G+G^2+G^3,即得答案 
*/
#include <iostream> 
#include <cstdio>
#include <cstring>
#include <map>

using namespace std;

#define MOD 2008

#define Mat 35 //矩阵大小
  
struct mat{//矩阵结构体，a表示内容，r行c列 矩阵从1开始  
    int a[Mat][Mat];
    int r, c;  
    mat() {  
        r = c = 0;  
        memset(a, 0, sizeof(a));  
    }  
};  

void print(mat m) {  
    //printf("%d\n", m.size);  
    for(int i = 0; i < m.r; i++) {  
        for(int j = 0; j < m.c; j++) printf("%d ", m.a[i][j]);  
        putchar('\n');  
    }  
}  
  
mat mul(mat m1, mat m2, int mod) {  
    mat ans  = mat();
    ans.r = m1.r, ans.c = m2.c;  
    for(int i = 1; i <= m1.r; i++)  
        for(int j = 1; j <= m2.r; j++)  
            if(m1.a[i][j])
                for(int k = 1; k <= m2.c; k++)  
                    ans.a[i][k] = (ans.a[i][k] + m1.a[i][j] * m2.a[j][k]) % mod;  
    return ans;  
}

mat add(mat m1, mat m2, int mod) {  
    mat ans  = mat();
    ans.r = ans.c = m1.r;  
    for(int i = 1; i <= m1.r; i++)  
        for(int j = 1; j <= m1.r; j++)  
            ans.a[i][j] = (m1.a[i][j] + m2.a[i][j]) % mod;  
    return ans;  
}    

mat quickmul(mat m, int n, int mod) {  
    mat ans = mat();  
    for(int i = 1; i <= m.r; i++) ans.a[i][i] = 1;  
    ans.r = m.r, ans.c = m.c;  
    while(n) {  
        if(n & 1) ans = mul(m, ans, mod);  
        m = mul(m, m, mod);  
        n >>= 1;  
    }  
    return ans;  
}  

mat A;

mat sum(mat m, int n, int mod) {   //m^1+m^2+...+m^n
    if(n == 1) return m;
    if(n & 1) return add(quickmul(m, n, mod), sum(m, n-1, mod), mod);             //是否加m^n 
    else return mul(add(quickmul(m, n>>1, mod), A, mod), sum(m, n>>1, mod), mod); // (1 + m^(n/2)) * (m + m^2 +...+ m^(n/2))
}

/* 
初始化ans矩阵 
mat ans = mat(); 
ans.r = R, ans.c = C;  
ans = quickmul(ans, n, mod); 
*/

int main() {
    A = mat();  
    for(int i = 1; i <= 30; i++) A.a[i][i] = 1;  
    int n;
    while(~scanf("%d", &n)) {
        mat G = mat();
        G.r = G.c = 30;
        map <int, int> mp;
        int rank = 1;
        while(n--) {
            int a, b;
            scanf("%d%d", &a, &b);
            if(!mp[a]) mp[a] = rank++;
            if(!mp[b]) mp[b] = rank++;
            G.a[mp[a]][mp[b]]++;
        }
        int k;
        scanf("%d", &k);
        for(int i = 0; i < k; i++) {
            int v1, v2, t1, t2;
            scanf("%d%d%d%d", &v1, &v2, &t1, &t2);
            v1 = mp[v1];
            v2 = mp[v2];
            if(t1 > t2) swap(t1, t2);
            if(!t1){
                if(!t2) puts("0");
                else {
                    int ans = sum(G, t2, MOD).a[v1][v2];
                    printf("%d\n", ans);
                }
            }
            else if(t1 == 1) {
                int ans = sum(G, t2, MOD).a[v1][v2];
                printf("%d\n", ans);
            }
            else {
                int ans1 = sum(G, t1-1, MOD).a[v1][v2];
                int ans2 = sum(G, t2, MOD).a[v1][v2];
                printf("%d\n", (ans2 - ans1 + MOD) % MOD);
            }
        }
    }
    return 0;
}