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题目链接:点击打开链接
题意:
给定n个点的树,染两种颜色,不同颜色不能相邻且要给尽可能多的节点染色。求颜色A和颜色B可能的染色节点个数。(copy from figo)
而且至少一个点染A,至少一个点染B
dp[i][j]=1表示i点染j个A色是可行的 =0表示不可行。
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.text.DecimalFormat;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Queue;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
public class Main {
ArrayList<Integer> L = new ArrayList(), R = new ArrayList();
int[][] dp = new int[N][N];//表示i点的子树 染j个红色时能得到最多几个蓝色
int[] siz = new int[N], ans = new int[N];
int n;
void dfs(int u, int fa){
for(int i = 2; i <= n; i++)dp[u][i] = 0;
dp[u][0] = 1;
siz[u] = 1;
for(int i = head[u]; i!=-1; i = edge[i].nex){
int v = edge[i].to; if(v == fa)continue;
dfs(v, u);
siz[u] += siz[v];
for(int j = n-1; j >= 0; j--)
if(dp[u][j]>0)
dp[u][j+siz[v]] = 1;
}
int s = n-siz[u];
for(int i = n-1; i >= 0; i--)
if(dp[u][i]>0)dp[u][i+s] = 1;
for(int i = 1; i < n-1; i++)
if(dp[u][i]>0)ans[i] = 1;
}
void work() throws Exception {
init_edge();
n = Int();
for(int i = 0; i <= n; i++)ans[i] = 0;
for(int i = 1, u, v; i< n; i++){
u =Int(); v = Int();
add(u,v,1); add(v,u,1);
}
dfs(1,-1);
int len = 0;
for(int i = 1; i < n-1; i++)len += ans[i];
out.println(len);
for(int i = 1; i < n-1; i++)
if(ans[i]>0){
out.println(i+" "+(n-1-i));
}
}
public static void main(String[] args) throws Exception {
Main wo = new Main();
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
// in = new BufferedReader(new InputStreamReader(new FileInputStream(new
// File("input.txt"))));
// out = new PrintWriter(new File("output.txt"));
wo.work();
out.close();
}
static int N = 5050;
static int M = N * 2;
DecimalFormat df = new DecimalFormat("0.0000");
static int inf = (int) 1e9;
static long inf64 = (long) 1e18;
static double eps = 1e-8;
static double Pi = Math.PI;
static int mod = (int) 1e9 + 7;
private String Next() throws Exception {
while (str == null || !str.hasMoreElements())
str = new StringTokenizer(in.readLine());
return str.nextToken();
}
private int Int() throws Exception {
return Integer.parseInt(Next());
}
private long Long() throws Exception {
return Long.parseLong(Next());
}
private double Double() throws Exception {
return Double.parseDouble(Next());
}
StringTokenizer str;
static Scanner cin = new Scanner(System.in);
static BufferedReader in;
static PrintWriter out;
class Edge{
int from, to, dis, nex;
Edge(){}
Edge(int from, int to, int dis, int nex)
{
this.from = from; this.to = to; this.dis = dis; this.nex = nex;
}
}
Edge[] edge = new Edge[M<<1];
int[] head = new int[N]; int edgenum;
void init_edge(){ for(int i = 0; i < N; i++)head[i] = -1; edgenum = 0;}
void add(int u, int v, int dis){
edge[edgenum] = new Edge(u, v, dis, head[u]);
head[u] = edgenum++;
}
/*
*/
int upper_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;
int pos = r;
r--;
while (l <= r) {
int mid = (l + r) >> 1;
if (A[mid] <= val) {
l = mid + 1;
} else {
pos = mid;
r = mid - 1;
}
}
return pos;
}
int lower_bound(int[] A, int l, int r, int val) {// upper_bound(A+l,A+r,val)-A;
int pos = r;
r--;
while (l <= r) {
int mid = (l + r) >> 1;
if (A[mid] < val) {
l = mid + 1;
} else {
pos = mid;
r = mid - 1;
}
}
return pos;
}
int Pow(int x, int y) {
int ans = 1;
while (y > 0) {
if ((y & 1) > 0)
ans *= x;
y >>= 1;
x = x * x;
}
return ans;
}
double Pow(double x, int y) {
double ans = 1;
while (y > 0) {
if ((y & 1) > 0)
ans *= x;
y >>= 1;
x = x * x;
}
return ans;
}
int Pow_Mod(int x, int y, int mod) {
int ans = 1;
while (y > 0) {
if ((y & 1) > 0)
ans *= x;
ans %= mod;
y >>= 1;
x = x * x;
x %= mod;
}
return ans;
}
long Pow(long x, long y) {
long ans = 1;
while (y > 0) {
if ((y & 1) > 0)
ans *= x;
y >>= 1;
x = x * x;
}
return ans;
}
long Pow_Mod(long x, long y, long mod) {
long ans = 1;
while (y > 0) {
if ((y & 1) > 0)
ans *= x;
ans %= mod;
y >>= 1;
x = x * x;
x %= mod;
}
return ans;
}
int gcd(int x, int y) {
if (x > y) {
int tmp = x;
x = y;
y = tmp;
}
while (x > 0) {
y %= x;
int tmp = x;
x = y;
y = tmp;
}
return y;
}
int max(int x, int y) {
return x > y ? x : y;
}
int min(int x, int y) {
return x < y ? x : y;
}
double max(double x, double y) {
return x > y ? x : y;
}
double min(double x, double y) {
return x < y ? x : y;
}
long max(long x, long y) {
return x > y ? x : y;
}
long min(long x, long y) {
return x < y ? x : y;
}
int abs(int x) {
return x > 0 ? x : -x;
}
double abs(double x) {
return x > 0 ? x : -x;
}
long abs(long x) {
return x > 0 ? x : -x;
}
boolean zero(double x) {
return abs(x) < eps;
}
double sin(double x) {
return Math.sin(x);
}
double cos(double x) {
return Math.cos(x);
}
double tan(double x) {
return Math.tan(x);
}
double sqrt(double x) {
return Math.sqrt(x);
}
}Codeforces 212E IT Restaurants 树形dp(水
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原文地址:http://blog.csdn.net/qq574857122/article/details/44515889