【hdu4903】The only survival

n个点的完全图，每条边的权值在1~L间，求有多少种图使得1到n的最短路长度为k。n ,k<= 12 L <= 10^9

暴力：枚举起点到每个点的最短路di，按di从小到大排序，求出方案。

我们并不关心具体点的di，枚举每个di的数量并统计。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define Rep(i, x, y) for (int i = x; i <= y; i ++)
#define Dwn(i, x, y) for (int i = x; i >= y; i --)
#define RepE(i, x) for(int i = pos[x]; i; i = g[i].nex)
using namespace std;
typedef long long LL;
const int N = 15, mod = 1000000007;
int T, n, L, m, l, d
; LL ans, u
, C

;
void Find(int x, LL y) {
LL n1 = 1, num = 1, n3, n2 = 1, t = 0, num0;
Rep(j, 1, l) n1 = n1 * (L - (x - d[j] - 1)) % mod, n2 = n2 * (L - (x - d[j])) % mod;
n3 = n1, n1 = (n1 - n2 + mod) % mod;
if (x == m) {
Rep(j, l + 1, n - 1) num = (num * n3 % mod) * u[j - l - 1] % mod;
(num *= u[n - l - 1]) %= mod;
(ans += (n1 * num % mod) * y) %= mod;
return ;
}
Find(x + 1, y);
while (l < n - 1) {
d[++ l] = x, num = (num * n1 % mod) * u[t] % mod; t ++; num0 = num;
if (l != n - 1) num0 = num0 * C[n - l + t - 1][t] % mod;
Find(x + 1, y * num0 % mod);
}
while (d[l] == x) l --;
}
int main()
{
scanf ("%d", &T);
C[0][0] = 1;
Rep(i, 1, 12) {
C[i][0] = 1;
Rep(j, 1, 12) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
}
while (T --) {
scanf ("%d%d%d", &n, &m, &L);
ans = 0; u[0] = 1;
Rep(i, 1, n) u[i] = u[i - 1] * L % mod;
d[l = 1] = 0; Find(1, 1);
printf("%I64d\n", ans);
}

return 0;
}