题目中文的不解释啊。

。。

须要注意的就是：离散数学中，有向图的邻接矩阵A表示全部点之间路径长度为1的路径数量，A^n则表示路径长度为n的路径数量。故须要求某两点在(A^t1)~(A^t2)的路径数量之和。

奥运

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2251    Accepted Submission(s): 572

   
   

    
    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
   
   

   
   

    
    0
1506
0
   
   

</pre><pre style="font-family: 'Courier New'; background-color: rgb(244, 251, 255);"><pre name="code" class="cpp">#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?
0:x)#define mod 2008const int maxn = 210;using namespace std;struct matrix{    int f[31][31];};matrix p[10001];map<int, int>mp;matrix mul(matrix a, matrix b, int n){    matrix c;    memset(c.f, 0, sizeof(c.f));    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            for(int k = 0; k < n; k++) c.f[i][j] += a.f[i][k]*b.f[k][j];            c.f[i][j] %= mod;        }    }    return c;}matrix pow_mod(matrix a, int b, int n){    matrix s;    memset(s.f, 0 , sizeof(s.f));    for(int i = 0; i < n; i++) s.f[i][i] = 1;    while(b)    {        if(b&1) s = mul(s, a, n);        a = mul(a, a, n);        b >>= 1;    }    return s;}matrix Add(matrix a,matrix b, int n)  {    matrix c;    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            c.f[i][j] = a.f[i][j]+b.f[i][j];            c.f[i][j] %= mod;        }    }    return c;}int main(){    int n, m;    while(scanf("%d",&n)!=EOF)    {        int u, v;        int ans = 0;        mp.clear();        memset(p[0].f, 0, sizeof(p[0].f));        for(int i = 0; i < n; i++)        {            scanf("%d %d",&u, &v);            if(mp.find(u) == mp.end()) mp[u] = ans++;            if(mp.find(v) == mp.end()) mp[v] = ans++;            p[0].f[mp[u]][mp[v]] ++;        }        for(int i = 1; i < 10001; i++) p[i] = mul(p[i-1], p[0], ans);        scanf("%d",&m);        int t1, t2, v1, v2;        while(m--)        {            scanf("%d %d %d %d",&v1, &v2, &t1, &t2);            if(t1 > t2) swap(t1,t2);            if(mp.find(v1) == mp.end() || mp.find(v2) == mp.end() || t1 == 0 && t2 == 0)            {                puts("0");                continue;            }            int sum = 0;            for(int i = t1-1; i < t2; i++)             {                if(i == -1) continue;                sum += p[i].f[mp[v1]][mp[v2]]%mod;            }            printf("%d\n",sum%mod);            ///cout<<(sum%mod)<<endl;        }    }    return 0;}