Codeforces Round 245 div1A&div2C Xor-tree dfs暴搜
2014-05-12 12:23
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题目链接点这儿
给一颗树,1为根,每次可以选一张牌执行一次翻转操作,除了自身外,自己的孙子(嘛,就是儿子的儿子啦),孙子的孙子都会翻转。问最少多少次翻转可以把题中给出的原始状态变成要的理想状态。
这题意,dfs肯定dfs啦因为翻转不会改变父亲们的状态,所以只要一点一点的向下dfs就好啦。没什么蛋疼的地方╮( ̄▽ ̄")╭
下面是代码
#include <bits/stdc++.h>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)>(b)?(b):(a))
#define rep(i,initial_n,end_n) for(int (i)=(initial_n);(i)<(end_n);i++)
#define repp(i,initial_n,end_n) for(int (i)=(initial_n);(i)<=(end_n);(i)++)
#define reep(i,initial_n,end_n) for((i)=(initial_n);(i)<(end_n);i++)
#define reepp(i,initial_n,end_n) for((i)=(initial_n);(i)<=(end_n);(i)++)
#define eps 1.0e-9
using namespace std;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef __int64 ll;
typedef unsigned __int64 ull;
const double pi = acos(-1);
int getint(){
int x=0,tmp=1; char c=getchar();
while((c<'0'||c>'9')&&c!='-') c=getchar();
if(c == '-') c=getchar() , tmp = -1;
while(c>='0'&&c<='9') x*=10,x+=(c-'0'),c=getchar();
return x*tmp;
}
vector<int> v[100010], ans;
int init[100010], need[100010], a[2] = { 0 };
//void dfs(int now, int time, int a[]);
void dfs(int now, int fa, int toplus, int next);
int main() {
int n, tmp, tmpp;
n = getint();
rep(i, 0, n-1) {
tmp = getint(), tmpp = getint();
v[tmp].push_back(tmpp);
v[tmpp].push_back(tmp);
}
repp(i, 1, n) init[i] = getint();
repp(i, 1, n) need[i] = getint();
//dfs(1, 0, a);
dfs(1, 0, 0, 0);
sort(ans.begin(), ans.end());
printf("%d\n", ans.size());
rep(i, 0, ans.size()) printf("%d\n", ans[i]);
return 0;
}
void dfs(int now, int fa, int toplus, int next) {
if((init[now] ^ need[now]) != toplus) {
toplus ^= 1;
ans.push_back(now);
}
int len = v[now].size();
rep(i, 0, len) if(v[now][i] != fa) dfs(v[now][i], now, next, toplus);
}
给一颗树,1为根,每次可以选一张牌执行一次翻转操作,除了自身外,自己的孙子(嘛,就是儿子的儿子啦),孙子的孙子都会翻转。问最少多少次翻转可以把题中给出的原始状态变成要的理想状态。
这题意,dfs肯定dfs啦因为翻转不会改变父亲们的状态,所以只要一点一点的向下dfs就好啦。没什么蛋疼的地方╮( ̄▽ ̄")╭
下面是代码
#include <bits/stdc++.h>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)>(b)?(b):(a))
#define rep(i,initial_n,end_n) for(int (i)=(initial_n);(i)<(end_n);i++)
#define repp(i,initial_n,end_n) for(int (i)=(initial_n);(i)<=(end_n);(i)++)
#define reep(i,initial_n,end_n) for((i)=(initial_n);(i)<(end_n);i++)
#define reepp(i,initial_n,end_n) for((i)=(initial_n);(i)<=(end_n);(i)++)
#define eps 1.0e-9
using namespace std;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef __int64 ll;
typedef unsigned __int64 ull;
const double pi = acos(-1);
int getint(){
int x=0,tmp=1; char c=getchar();
while((c<'0'||c>'9')&&c!='-') c=getchar();
if(c == '-') c=getchar() , tmp = -1;
while(c>='0'&&c<='9') x*=10,x+=(c-'0'),c=getchar();
return x*tmp;
}
vector<int> v[100010], ans;
int init[100010], need[100010], a[2] = { 0 };
//void dfs(int now, int time, int a[]);
void dfs(int now, int fa, int toplus, int next);
int main() {
int n, tmp, tmpp;
n = getint();
rep(i, 0, n-1) {
tmp = getint(), tmpp = getint();
v[tmp].push_back(tmpp);
v[tmpp].push_back(tmp);
}
repp(i, 1, n) init[i] = getint();
repp(i, 1, n) need[i] = getint();
//dfs(1, 0, a);
dfs(1, 0, 0, 0);
sort(ans.begin(), ans.end());
printf("%d\n", ans.size());
rep(i, 0, ans.size()) printf("%d\n", ans[i]);
return 0;
}
void dfs(int now, int fa, int toplus, int next) {
if((init[now] ^ need[now]) != toplus) {
toplus ^= 1;
ans.push_back(now);
}
int len = v[now].size();
rep(i, 0, len) if(v[now][i] != fa) dfs(v[now][i], now, next, toplus);
}
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