[] [NOIP2016 improve JZOJ4816 A group of five school entrance exam 4] label

Title effect: There are a \ (n-\) tree points, each point you give assign a \ ([1, m] \ ) integer values within the weight, so that two adjacent right-point value a difference of at least \ (k \) , ask a few programs.
\ (n, k \ leq 100 , m \ leq 10 ^ 9 \)

Solution

Set \ (f [i] [j ] \) represents \ (I \) subtree rooted satisfies a condition, and \ (I \) weight value \ (J \) of the program number, and transfer with prefix optimization .
Can be found by playing table \ (f [i] [j ] \) is symmetrical, ie \ (F [I] [J] = F [I] [m-J +. 1] \) , and left up to \ ((n-1) * k \) different numbers, the middle section is the same. So we recorded before \ ((n-1) * k \) states, and \ (pos [i] \) represents a start from which are identical, will be able to \ (O (1) \) is determined and the prefix.

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long ll;
const int N=107,M=10017;
const ll P=1000000007;

int T,n,m,k;
ll f[N][M],s[N][M],pos[N];

int tot,st[N],to[N<<1],nx[N<<1];
void add(int u,int v){to[++tot]=v,nx[tot]=st[u],st[u]=tot;}

ll sum(int u,int j){
    ll ret=0;
    if(j<=10010)return s[u][j];
    else if(j<pos[u])return s[u][j];
    else{
        ret=(ret+s[u][pos[u]-1])%P;
        if(j<=m-pos[u]+1)ret=(ret+f[u][pos[u]]*(j-pos[u]+1)%P)%P;
        else{
            ret=(ret+f[u][pos[u]]*(m-2*pos[u]+2)%P)%P;
            ret=(ret+s[u][pos[u]-1]-s[u][m-j]+P)%P;
        }
        return ret;
    }
}

void dfs(int u,int from){
    for(int i=st[u];i;i=nx[i])if(to[i]!=from)dfs(to[i],u);
    for(int j=1;j<=10010&&j<=m;++j){
        f[u][j]=1;
        for(int i=st[u];i;i=nx[i])if(to[i]!=from){
            if(k==0)f[u][j]=f[u][j]*sum(to[i],m)%P;
            else{
                ll tmp=0;
                if(j-k>0)tmp=(tmp+sum(to[i],j-k))%P;
                if(j+k<=m)tmp=(tmp+sum(to[i],m)-sum(to[i],j+k-1)+P)%P;
                f[u][j]=f[u][j]*tmp%P;
            }
        }
    }
    for(int j=1;j<=10010&&j<=m;++j)if(f[u][j]==f[u][j+1]){pos[u]=j;break;}
    for(int j=1;j<=10010&&j<=m;++j)s[u][j]=(s[u][j-1]+f[u][j])%P;
}

int main(){
    freopen("label.in","r",stdin);
    //freopen("label.out","w",stdout);
    scanf("%d",&T);
    while(T--){
        tot=0;
        memset(st,0,sizeof(st));
        memset(f,0,sizeof(f));
        memset(s,0,sizeof(s));
        memset(pos,0,sizeof(pos));
        memset(to,0,sizeof(to));
        memset(nx,0,sizeof(nx));
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1,u,v;i<n;++i)scanf("%d%d",&u,&v),add(u,v),add(v,u);
        dfs(1,0);
        printf("%lld\n",sum(1,m));
    }
    return 0;
}