POJ1094 Sorting It All Out（拓扑排序）

Sorting It All Out

• 总时间限制: 1000ms
• 内存限制: 65536kB

描述

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 problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.

输入

Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.

输出

For each problem instance, output consists of one line. This line should be one of the following three:

Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.

where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence.

样例输入

4 6
A<B
A<C
B<C
C<D
B<D
A<B
3 2
A<B
B<A
26 1
A<Z
0 0

样例输出

Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.

来源

East Central North America 2001

题意

• 如果能确定有序序列，输出最早能够确定顺序的关系式的数量以及该序列
• 如果出现冲突，输出最早能够确定冲突的关系式的数量
• 如果不能确定也不存在冲突，输出无法得到

代码

StatusAccepted
Time1ms
Memory128kB
Length1779
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 30;
int indegree[maxn], _indegree[maxn];
int n, m;
vector <int> E[maxn];
char ans[maxn];
queue <int> Q;

void init()
{
for (int i = 0; i < n; ++i)
{
E[i].clear();
indegree[i] = 0;
}
}

int Topsort()
{
int size = 0, ret = 1;
memcpy(_indegree, indegree, sizeof(_indegree));
for (int i = 0; i < n; ++i)
{
if (_indegree[i] == 0)
Q.push(i);
}
while (!Q.empty())
{
if (Q.size() > 1) // 当前状态下有多个入度为0的节点，选择的顺序不能确定
ret = 0;
int u = Q.front();
Q.pop();
ans[size++] = char(u + 'A');
for (int i = 0; i < E[u].size(); ++i)
{
int v = E[u][i];
if ((--_indegree[v]) == 0)
Q.push(v);
}
}
if (size < n) // 出现环
return -1;
//    for (int i = 0; i < n; ++i) // 判断环，同上
//        if (_indegree[i] != 0)
//            return -1;
ans[size] = '\0'; // 不加会报错
return ret;
}

int main()
{
char s[10];
int u, v;
while (scanf("%d%d", &n, &m) != EOF && n && m)
{
init();
int flag = 0;
for (int i = 0; i < m; ++i)
{
scanf("%s", s);
if (flag) // 如果已经判断完成只需继续输入即可
continue;
u = s[0] - 'A', v = s[2] - 'A';
E[u].push_back(v);
indegree[v]++;
flag = Topsort();
if (flag == 1)
printf("Sorted sequence determined after %d relations: %s.\n", i + 1, ans);
else if (flag == -1)
printf("Inconsistency found after %d relations.\n", i + 1);
}
if (!flag)
printf("Sorted sequence cannot be determined.\n");
}
return 0;
}

The end.
2018-07-20 星期五

模板

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
const int maxn = 1010;
int vis[maxn];
vector <int> E[maxn];
vector <int> V;
int n, m;

bool dfs(int u)
{
vis[u] = -1; // 当前
for (int i = 0; i < E[u].size(); ++i)
{
int v = E[u][i];
if (vis[v] == -1) // 出现环
return false;
else if (!vis[v])
dfs(v);
}
V.push_back(u);
vis[u] = 1;
return true;
}

bool TopSort()
{
for (int i = 1; i <= n; ++i)
if (!vis[i])
{
if (!dfs(i))
return false;
}
return true;
}

int main()
{

scanf("%d%d", &n, &m);
int u, v;
for (int i = 0; i < m; ++i)
{
scanf("%d%d", &u, &v);
E[u].push_back(v);
}
if (TopSort())
{
for (int i = V.size() - 1; i >= 0; --i)
printf("%d ", V[i]);
}
else
printf("No\n");
}

/*
/Users/taifu/Codes/CPP/cmake-build-debug/CPP
13 15
1 2
1 6
1 7
3 1
3 4
4 6
6 5
7 4
7 10
8 7
9 8
10 11
10 12
10 13
12 13
9 8 3 1 7 10 12 13 11 4 6 5 2
Process finished with exit code 0
*/