UVa 10129 - Play on Words

来源：互联网发布：淘宝店技巧编辑：程序博客网时间：2024/06/09 16:23

題目：考古學家在迷宮裡發現了一些單詞，把他們首尾相接拼在一起，放到門的機關上，

如果連接的字母相同，們就會打開，現在給你單詞，問門是否能打開。

分析：圖論，歐拉路徑，并查集。利用并查集求解歐拉路徑即可。

歐拉路徑存在條件：

1.存在邊的節點聯通；

2.所有節點的入度和出度相同，或者有一個點入度比出度大1及另一個點出度比入度大1；

說明：╮(╯▽╰)╭。

#include <cstdlib>#include <cstring>#include <cstdio>#define NODE_SIZE 26#define EDGE_SIZE 100001int in[NODE_SIZE];int out[NODE_SIZE];typedef struct _edge{int point1;int point2;}edge;edge E[EDGE_SIZE];//union_setint sets[NODE_SIZE];int rank[NODE_SIZE];void set_inital(int a, int b){for (int i = a; i <= b; ++ i) {rank[i] = 0;sets[i] = i;}}int set_find(int a){if (a != sets[a])sets[a] = set_find(sets[a]);return sets[a];}void set_union(int a, int b){if (rank[a] < rank[b])sets[a] = b;else {if (rank[a] == rank[b])rank[a] ++;sets[b] = a;}}//end_union_setint euler(int n, int m){memset(in, 0, sizeof(in));memset(out, 0, sizeof(out));set_inital(0, n);for (int i = 0; i < m; ++ i) {int A = set_find(E[i].point1);int B = set_find(E[i].point2);if (A != B) set_union(A, B);out[E[i].point1] ++;in[E[i].point2] ++;}int ans = 0, big = 0, small = 0, flag = 0;for (int i = 0; i < n; ++ i) {if (in[i] | out[i]) ans += (sets[i]==i);big += (in[i] == out[i]+1);small += (in[i] == out[i]-1);if (in[i] > out[i]+1 || out[i] > in[i]+1) return 0;}return (ans == 1 && ((big == 0 && small == 0) || (big == 1 && small == 1)));}int main(){int  n, m;char buf[1001];while (~scanf("%d",&n)) while (n --) {scanf("%d",&m);for (int i = 0; i < m; ++ i) {scanf("%s",buf);E[i].point1 = buf[0]-'a';E[i].point2 = buf[strlen(buf)-1]-'a';}if (euler(26, m))printf("Ordering is possible.\n");else printf("The door cannot be opened.\n");}return 0;}

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