Codeforces Round #405 D. Bear and Tree Jumps 树形DP

传送门

题意：

你可以从树上的节点一次最多走k条边。 
(称为跳一次); 
树为无权树; 
然后问你任意两点之间的条的次数的和为多少; 

做法：

枚举边，看这条边左边的点的数目和右边的点的数目分别为多少->num1和num2 

num1*num2就是经过这条边的路径个数; 
累加所有的边的这个值就是任意两点之间的距离了; 

dp[i][j]表示离i这个点，距离%k为j的结点数。

很自然得到ceil(L/k)= L/k+sigma_(i=0)-(i=k-1)/k;

因为枚举的是x-G[x][i]这条边，所以每次处理的时候，最后dp[x][j]+=dp[i][j]。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=4e5+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;LL dp[maxn][6],a[maxn];int n,k;vector<int>G[maxn];LL ans=0;void dfs(int x,int fa,int deep){    a[x]=1;    dp[x][deep%k]=1;//    for(int i:G[x])    {        if(i==fa)            continue;        dfs(i,x,deep+1);        for(int j=0;j<k;j++)        {            for(int z=0;z<k;z++)            {                int len=(j+z-2*deep)%k;                int rev=(k-len)%k;                ans+=rev*dp[x][j]*dp[i][z];            }        }        a[x]+=a[i];        for(int j=0;j<k;j++)            dp[x][j]+=dp[i][j];//        ans+=(n-a[i])*a[i];//    }}void solve(){    W(s_2(n,k)!=EOF)    {        FOr(1,n,i)        {            int u,v;            s_2(u,v);            G[u].pb(v);            G[v].pb(u);        }        dfs(1,-1,0);        print(ans/k);    }}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    //s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case %d:\n",cas);        solve();    }}