题目大意:有N个木棍,每根木棍两端被涂上颜色,现在给定每个木棍两端的颜色,不同木棍之间拼接需要颜色相同的
端才可以,问最后能否将N个木棍拼接在一起。
解题思路:欧拉通路+并查集+字典树。欧拉通路,每个节点的统计度,度为奇数的点不能超过2个。并查集,判断节点
是否完全联通。字典树,映射颜色。
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 5000005;
const int sigma_size = 26;
typedef pair<int,int> pii;
int N, E, far[maxn], cnt[maxn];
pii ed[maxn];
struct Tire {
int sz, g[maxn][sigma_size];
int val[maxn];
void init();
int idx(char ch);
int insert(char* s);
}T;
inline int getfar(int x) {
return x == far[x] ? x : far[x] = getfar(far[x]);
}
bool judge() {
int ret = 0;
for (int i = 1; i <= N; i++) {
if (cnt[i]&1)
ret++;
}
if (ret > 2)
return false;
for (int i = 0; i < E; i++) {
int p = getfar(ed[i].first);
int q = getfar(ed[i].second);
if (p != q) {
N--;
far[p] = q;
}
}
return N <= 1;
}
int main () {
T.init();
N = E = 0;
char a[11], b[11];
while (scanf("%s%s", a, b) == 2) {
int p = T.insert(a);
int q = T.insert(b);
cnt[p]++, cnt[q]++;
ed[E].first = p;
ed[E].second = q;
E++;
}
printf("%s\n", judge() ? "Possible" : "Impossible");
return 0;
}
void Tire::init() {
sz = 1;
val[0] = 0;
memset(g[0], 0, sizeof(g[0]));
}
int Tire::idx(char ch) {
return ch - ‘a‘;
}
int Tire::insert(char* s) {
int u = 0, n = strlen(s);
for (int i = 0; i < n; i++) {
int v = idx(s[i]);
if (g[u][v] == 0) {
g[u][v] = sz;
memset(g[sz], 0, sizeof(g[sz]));
val[sz++] = 0;
}
u = g[u][v];
}
if (val[u] == 0) {
val[u] = ++N;
far[N] = N;
cnt[N] = 0;
}
return val[u];
}
时间: 2024-10-16 08:49:52