nyoj Sorting It All Out (拓扑排序)

三种情况分别是:

1. 在某位置可以确定拓扑排序。

2. 在某位置出现了环

3. 到最后都不能确定拓扑排序(某一位置入度为0的点有多个),可以续输入执行下去。

每输入一组数据都要做一次判断

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<vector>
 4 #include<queue>
 5 using namespace std;
 6 const int N = 105;
 7 int n, m, in[N], temp[N], ans[N], t, pos, num;
 8 char X, O, Y;
 9 vector<int>G[N];
10 queue<int>q;
11
12 void init(){
13     memset(in, 0, sizeof(in));
14     for (int i = 0; i <= n; ++i){
15         G[i].clear();
16     }
17 }
18
19 int topoSort(){
20     while (!q.empty())q.pop();
21     for (int i = 0; i<n; ++i)if (in[i] == 0){
22         q.push(i);
23     }
24     pos = 0;
25     bool unSure = false;
26     while (!q.empty()){
27         if (q.size()>1) unSure = true;
28         int t = q.front();
29         q.pop();
30         ans[pos++] = t;
31         for (int i = 0; i<G[t].size(); ++i){
32             if (--in[G[t][i]] == 0)
33                 q.push(G[t][i]);
34         }
35     }
36     if (pos<n) return 1;
37     if (unSure)  return 2;
38     return 3;
39 }
40
41 int main(){
42     int x, y, i, flag, ok, stop;
43     while (~scanf("%d%d%*c", &n, &m),n+m){
44         init();
45         flag = 2;
46         ok = false;
47         for (i = 1; i <= m; ++i){
48             scanf("%c%c%c%*c", &X, &O, &Y);
49             if (ok) continue;
50             x = X - ‘A‘, y = Y - ‘A‘;
51             if (O == ‘<‘){
52                 G[y].push_back(x);
53                 ++in[x];
54             }
55             else if (O == ‘>‘){
56                 G[x].push_back(y);
57                 ++in[y];
58             }
59             memcpy(temp, in, sizeof(in));
60             flag = topoSort();
61             memcpy(in, temp, sizeof(temp));
62             if (flag != 2){
63                 stop = i;
64                 ok = true;
65             }
66         }
67         if (flag == 3){
68             printf("Sorted sequence determined after %d relations: ", stop);
69             for (int i = pos - 1; i >= 0; --i)
70                 printf("%c", ans[i] + ‘A‘);
71             printf(".\n");
72         }
73         else if (flag == 1){
74             printf("Inconsistency found after %d relations.\n", stop);
75         }
76         else{
77             printf("Sorted sequence cannot be determined.\n");
78         }
79     }
80     return 0;
81 }

代码君

时间: 2024-08-28 13:45:51

nyoj Sorting It All Out (拓扑排序)的相关文章

Sorting It All Out 拓扑排序+确定点

这一道题的话  数据有一点问题    ........     例如 不过 还是   能理解一下  试试吧  ......... 3 5 A<B B<C C<A A<C B<A 这几组数据 明显反映出来 这是成环的 , 然后 按照 输入输出案例来说 这个是 有序的 ABC 题目要求     在每组数据的   第一行  给你需要排序 的 字母数    和  他们之间的关系数量      然后  输入每组数据    你首先许亚萍判断在输入  第几组 数据的时候 出现了 环    

nyoj349 poj1094 Sorting It All Out(拓扑排序)

nyoj349   http://acm.nyist.net/JudgeOnline/problem.php?pid=349poj1094   http://poj.org/problem?id=1094这两个题是一样的,不过在poj上A了才算真的过,ny上数据有一点弱. 题目大意输入n,m. 一共有n个字母(从A开始), m行语句每个语句“x﹤y”,说明x,y之间的偏序关系.让你判断是否可以通过这些关系得到一个唯一的升序序列,若能则输出这个序列并指出通过前多少条语句得出的,如果n个字母间存在矛

POJ 1094 Sorting It All Out 拓扑排序

Sorting It All Out Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to

POJ1094 Sorting It All Out 拓扑排序(深搜)

题意:给定n和m,n表示大写字母的前n个,m表示m个关系对,求能否确定唯一排序. 分析:分三种情况 (1)当输入完当前的关系对之后,已经能确定矛盾(不需要完全输入m个关系对时) eg. 3 3       A<B       B<A         B<C       当输入到第二对已经能判断矛盾,此时记下当前的所需要用到的关系对数ans=2:       接着输入,但是后面的关系对不作处理 (2) 当输入完当前的关系对之后,已经能确定唯一排序(不需要完全输入m个关系对,    但是必须

zoj 1060 Sorting It All Out(拓扑排序)

#include<bits/stdc++.h> using namespace std; vector< vector<char> >v; int n,m,use[30],cnt[30]; char ans[30],s[10]; int topsort(int x) { int i,j,t[30],f=1,r,c; memset(ans,0,sizeof(ans)); for(i=0; i<30; i++) t[i]=cnt[i]; c=0; r=0; while

Sorting It All Out (拓扑排序+思维)

An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this pro

[ACM] POJ 1094 Sorting It All Out (拓扑排序)

Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26801   Accepted: 9248 Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from sm

POJ 1094 Sorting It All Out【floyd传递闭包+拓扑排序】

Sorting It All Out Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 31994 Accepted: 11117 Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from small

hdu 5195 DZY Loves Topological Sorting BestCoder Round #35 1002 [ 拓扑排序 + 优先队列 || 线段树 ]

传送门 DZY Loves Topological Sorting Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 52 Problem Description A topological sort or topological ordering of a directed