2016.4.06test:problem2:bzoj: Search + Combinatorial Math _oi

The main idea: there is a complete picture of n points, each side has a 1~l weight, bear children want to know how many such graphs, so that from 1 to n the shortest short-circuit for k, because such a diagram may be a lot of you just output the number of solutions to 1e9+7 after the result can be. N,k<=12,l<=10^9

Analysis: N,k is very small, l large, considering the enumeration, enumerating the number of points with the shortest short-circuit as I a[i]

For two points i,j if DI=DJ, then this side of the Benquan is arbitrary, if DI<DJ, then the edge of the right can not be less than dj-di, so as to ensure that the shortest path is not changed, then the edge of the tmp1= (L (di-dj) +1) species may, which DI>=DJ

However, it is found that the shortest circuit property of Di must have at least one J to guarantee Dj+w (J,i) =di.

Because the Benquan W (j,i) of the Dj+w (j,i) >di has tmp2= (L (di-dj)) species may

Therefore, TMP1 and TMP2 respectively after the TMP1-TMP2 is to ensure the number of the shortest short-circuit nature of Di

Note that if di>=k, you don't have to control tmp2.

Finally, because we are enumerating the number of points with the shortest path to I, there is no order, so multiply it by (n-2).

But found that the same point in the same I will repeat, so there is a factorial divided by each a[i]

#include <cmath> #include <cstdio> #include <cstdlib> #include <iostream> #include <
Algorithm> using namespace std;
int a[15],b[15],n,k,l,t,cnt,bit[15],tt;
Long ans;
const int mod=1000000007;
		void Dfs (int x,int y) {if (x==k+1) {if (y+2!=n&&k==l) return;
        cnt=0;
        for (int i=1;i<=n;i++) for (int j=1;j<=a[i];j++) b[++cnt]=i;
        B[++cnt]=k;
        for (int i=1;i<=n-y-2;i++) b[++cnt]=k+1;
        Long Long ret=1;
            for (int i=1;i<=cnt;i++) {long long tmp1=1,tmp2=1; for (int j=0;j<i;j++) if (B[j]<b[i]) tmp1=tmp1* (L (b[i]-b[j)) +1%mod, tmp2=tmp2* (L (b[i]-b[j))%m
            Od
			if (b[i]<=k) ret=ret* (tmp1-tmp2+mod)%mod;
            else Ret=ret*tmp1%mod;
        for (int j=0;j<i;j++) if (b[j]==b[i]) ret=ret*l%mod;
		int q=tt;
		for (int i=1;i<x;i++) q/=bit[a[i]];
        Q/=BIT[N-Y-2];
        Ans= (ans+ret*q)%mod;
 Return   for (int i=0; i<=n-2-y; i++) {a[x]=i;
    DFS (X+1,y+i);
    int main () {bit[0]=1;
    for (int i=1; i<=12; i++) bit[i]=bit[i-1]*i;
    scanf ("%d%d%d", &n,&k,&l);
    if (l<k) {puts ("0"); return 0;}
    if (n==2) {puts ("1"); return 0;} Tt=1;
    for (int i=2;i<n-1;i++) tt*=i;
    DFS (1,0);
cout<<ans<<endl;
 }