标签:eof efi break secure lse iostream front 题意 return
题意:给你1e5个字符串,若前一个的末尾字母等于当前的首字母,则可以连在一起(成语接龙一个意思)判断是否可以将他们连在一起
题解:将首位看作点,单词看作边。变成欧拉回路问题。
判断出入度是否相等,再用并查集判一下连通性
(dfs/bfs也行:随便取一个点,搜索一遍。如果每个点都被标记,则是连通的。)
ac代码:
#define _CRT_SECURE_NO_WARNINGS #include "stdio.h" #include<stdio.h> #include<algorithm> #include<string> #include<vector> #include<list> #include<set> #include<iostream> #include<string.h> #include<queue> #include<string> #include<sstream> #include<stack> #include<cmath> using namespace std; const int maxn = 1e5 + 5; char s[maxn][1005]; char ss[1005]; int in[256], out[256]; //两个点入度不等于出度且一个差+1,一个-1 int f[256]; int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); } void un(int x, int y) { int u = find(x); int v = find(y); f[u] = v; } bool same(int x,int y) { return find(x) == find(y); } int main() { int t; cin >> t; while (t--) { int n; cin >> n; for (int i = 0; i < 256; i++) f[i] = i; memset(in, 0, sizeof(in)); memset(out, 0, sizeof(out)); for (int i = 0; i < n; i++) { scanf("%s", ss); int len = strlen(ss); out[ss[0]]++; in[ss[len - 1]]++; un(ss[0], ss[len - 1]); } int ok = 1; //连通 int yes = 0;//出入度 vector<int> temp; for (int i = ‘a‘; i <= ‘z‘; i++)if(out[i]+in[i]) {//出现过的字母 if (!same(‘a‘, i)) { ok = 0; break; } if (out[i] != in[i]) { temp.push_back(i); } } if (temp.size() == 2) { int f = temp.front(), b = temp.back(); if ((abs(in[f] - out[f]) == 1) && (abs(in[b] - out[b]) == 1) && (in[f] + in[b] == out[f] + out[b]))yes=1; } else if (temp.size() == 0)yes = 1; if (ok&&yes)cout << "Ordering is possible.\n"; else cout << "The door cannot be opened.\n"; } system("pause"); return 0; }
【紫书】Play on Words UVA - 10129 欧拉回路
标签:eof efi break secure lse iostream front 题意 return
原文地址:https://www.cnblogs.com/SuuT/p/8822871.html