Hdu4903 The only survival

The only survival

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 255    Accepted Submission(s): 120

Problem Description

There is an old country and the king fell in love with a devil. The devil always ask the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.
Something bad actually happen. The devil makes this kingdom's people infected by a disease called lolicon. Lolicon will take away people's life in silence. 
Although z*p is died, his friend, y*wan is not a lolicon. Y*wan is the only one in the country who is immune of lolicon, because he like the adult one so much. 
As this country is going to hell, y*wan want to save this country from lolicon, so he starts his journey.
You heard about it and want to help y*wan, but y*wan questioned your IQ, and give you a question, so you should solve it to prove your IQ is high enough.
The problem is about counting. How many undirected graphs satisfied the following constraints?
1. This graph is a complete graph of size n. 
2. Every edge has integer cost from 1 to L.
3. The cost of the shortest path from 1 to n is k.
Can you solve it?
output the answer modulo 10^9+7

Input

The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains 3 integers n,k,L.
T<=5 n,k<=12,L<=10^9.

Output

For each test case, output the answer in one line.

Sample Input

2 3 3 3 4 4 4

Sample Output

8 668

Author

WJMZBMR

Source

2014 Multi-University Training Contest 4

题目大意：有一张n个点的无向完全图，第i个点的编号是i，每条边的边权在1到L之间的正整数，问存在多少个图使得1到n的最短路是k。

分析：神题!

　　　一个暴力的做法是枚举所有点到1号点的最短路，记作d[i],如果d[i] == d[j],那么i,j之间的边的权值可以任意选，否则边权不能小于|di-dj|，且必然有一条边使得di + dist[i][j] = dj.只是枚举的复杂度高达13^12,承受不了.

　　　其实我们不需要考虑每个点的最短路具体是什么，只需要考虑数量就可以了.计数题的一个转换思路.那么在dfs的时候枚举每个最短路对应的点有多少个就可以了.然后就是极其琐碎的组合数运算.

　　　需要注意的是起点和终点的最短路都已经确定了.选取i个点的组合数要排除这两个点. 同时可能会有大于k的最短路出现，统统放到k+1处理.关于具体的组合数的运算我写在了代码里：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
const int mod = 1e9+7;
int T;
ll n,k,l,cnt[20],ans,c[25][25];

void init()
{
    c[0][0] = 1;
    for (int i = 1; i <= 20; i++)
    {
        c[i][0] = 1;
        for (int j = 1; j <= 20; j++)
            c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
}

ll qpow(ll a,ll b)
{
    ll res = 1;
    while (b)
    {
        if (b & 1)
            res = (res * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return res;
}

ll solve(ll x)
{
    if (!cnt[x])
        return 1;
    ll x1 = 1,x2 = 1;
    for (int i = 0; i < x; i++)
    {
        if (!cnt[i])
            continue;
        if (x - i > l)  //方案不合法，最短路会被更新
            return 0;
        x1 = (x1 * qpow(l - (x - i) + 1,cnt[i])) % mod;   //x1是所有能选的方案
        x2 = (x2 * qpow(l - (x - i),cnt[i])) % mod;  //x2是所有不合法的方案
    }
    if (x == k + 1)   //如果最短路大于k了，那么所有能选的方案都是合法的，因为只是把它们归到k+1这一类,最短路并不一定要等于k+1
        return qpow(x1,cnt[x]) % mod;
    x1 -= x2;
    if (x1 < 0)
        x1 += mod;
    return qpow(x1,cnt[x]) % mod;   //之前的方案数都是对于所有的i，1个x来计算的.
}

void dfs(ll dep,ll fangan,ll tot)  //tot传的是1
{
    if (dep == k)
    {
        for (int i = 1; i + tot <= n; i++)
        {
            ll temp = fangan * c[n - tot - 1][i - 1] % mod;   //剩下的点中选i-1个最短路为k的点，第i个点为终点，不考虑.
            temp = (temp * qpow(l,c[i][2])) % mod;   //这两行就是两两最短路相等的算方案数
            temp = (temp * qpow(l,c[n - tot - i][2])) % mod;
            cnt[k] = i;
            cnt[k + 1] = n - tot - i;
            temp = (temp * solve(k)) % mod;
            temp = (temp * solve(k + 1)) % mod;  //最短路大于k的都放到k+1处计算.
            ans = (ans + temp) % mod;
        }
        return;
    }
    for (int i = 0; i + tot < n; i++)
    {
        cnt[dep] = i;
        ll temp = fangan * qpow(l,c[i][2]) % mod;   //上面说的di == dj的情况，边权随便取
        temp = (temp * c[n - tot - 1][i]) % mod;   //能选的点中选i个点的方案数
        temp = (temp * solve(dep)) % mod;
        dfs(dep + 1,temp,i + tot);
    }
}

int main()
{
    scanf("%d",&T);
    init();
    while (T--)
    {
        scanf("%lld%lld%lld",&n,&k,&l);
        memset(cnt,0,sizeof(cnt));
        cnt[0] = 1;  //起点被确定了
        ans = 0;
        dfs(1,1,1);
        printf("%lld
",ans);
    }

    return 0;
}