E. Dasha and Puzzle 数学题

http://codeforces.com/contest/761/problem/E

给出一颗树，要求在坐标系中用平行于坐标轴的线描绘出来。

要求边不能相交，而且点的坐标唯一。

注意到2^1 + 2^2 + ..... + 2^n = 2^（n + 1） - 1

那就是说，如果第一条边的边长是2^（n + 1），那么后面的边选2^n 、 2^(n - 1)等等，相加起来也不会越过第一条，所以也就不会相交。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 30 + 20;
struct node {
int u, v, tonext;
} e[maxn * 2];
int num;
int first[maxn];
void add(int u, int v) {
++num;
e[num].u = u;
e[num].v = v;
e[num].tonext = first[u];
first[u] = num;
}
struct coor {
LL x, y;
coor(LL xx, LL yy) : x(xx), y(yy) {}
bool operator < (const struct coor & rhs) const {
if (x != rhs.x) return x < rhs.x;
else return y < rhs.y;
}
};
vector<struct coor>ans[maxn];
int tonext[4][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}};
bool vis[maxn];
const LL one = 1;
int haha[4];
void dfs(int cur, LL x, LL y, int son, int pre) {
ans[cur].push_back(coor(x, y));
//    aler.insert(coor(x, y));
vis[cur] = true;
int to = 0;
for (int i = first[cur]; i; i = e[i].tonext) {
int v = e[i].v;
if (vis[v]) continue;
if (to == pre) to++;
if (to == 4) to = 0;
dfs(v, x + tonext[to][0] * (one << son), y + tonext[to][1] * (one << son), son - 1, haha[to]);
to++;
}
}
void work() {
haha[0] = 2;
haha[1] = 3;
haha[2] = 0;
haha[3] = 1;
int n;
scanf("%d", &n);
for (int i = 1; i <= n - 1; ++i) {
int u, v;
scanf("%d%d", &u, &v);
add(u, v);
add(v, u);
}
for (int i = 1; i <= n; ++i) {
int t = -1;
for (int j = first[i]; j; j = e[j].tonext) {
t++;
}
if (t >= 4) {
//            cout << t << endl;
cout << "NO" << endl;
return;
}
}
//    LL y = 1 << 30;
//    y <<= 2;
//    cout << y << endl;
dfs(1, 1, 1, 32, -1);
cout << "YES" << endl;
for (int i = 1; i <= n; ++i) {
//        if (ans[i].size() == 0) {
//            cout << i << "ffff" << endl;
//            continue;
//        }
cout << ans[i][0].x << " " << ans[i][0].y << endl;
}
}

int main() {
#ifdef local
freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}

View Code