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Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 38356 | Accepted: 13525 |
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 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.
Input
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.
Output
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.
Sample Input
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
Sample Output
Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.题目大意：给出以下关系，①处理到第几个时能确定所有的关系，②处理到第几个时能发现矛盾，③没有矛盾所有点都能确定
思路：先是想到并查集，但是谁比谁大不好比较。后来构建有向图，每读入一组关系，就拓扑排序一次。
AC代码：
#include<iostream>
#include<cstdio>
#include<set>
#include<queue>
#include<string.h>
using namespace std;
int n, m, cnt;
int indegree[1010], temp[1010];
vector<int> G[1010];//邻接表
queue<int> q;
char ans[31];
bool check(int x, int y) {
for(int i = 0; i < G[x].size(); i++) {
if(G[x][i] == y)
return true;
}
return false;
}
int topo() {
for(int i = 0; i < n; i++) {//这里是跑所有的点
if(!indegree[i])
q.push(i);
}
bool unsure = false;//不确定的情况
cnt = 0;
while(!q.empty()) {//因为是跑所有的点，一旦出现有的点没有涉及，入读为0的点必然大于1
if(q.size() > 1)
unsure = true;
int u = q.front(); //取队列首个顶点u
ans[cnt++] = u + 'A';
q.pop();
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
indegree[v]--; //入度减一
if(indegree[v] == 0) {//入度为0，入队
q.push(v);
}
}
}
if(cnt < n)//有环，矛盾
return 2;
if(unsure)
return 3;
return 1;
}
char s[3];
int main() {
int step, flag;
while(~scanf("%d %d", &n, &m)) {
memset(G, 0, sizeof(G));
while(!q.empty()) q.pop();
if(n == 0 && m == 0)
break;
bool ok = false;
memset(indegree, 0, sizeof(indegree));
for(int i = 0; i < m; i++) {
scanf("%s", s);
if(ok)//就算知道了结果，也要读入所有数据
continue;
int l = s[0]-'A';
int r = s[2]-'A';
if(!check(l, r)) {//有过了，就不用再加入了
G[l].push_back(r);
indegree[r]++;
}
memcpy(temp, indegree, sizeof(indegree));//把当前入度情况记录下来
flag = topo();//拓扑判断一下
memcpy(indegree, temp, sizeof(temp));//再把入读情况放进入，进行下一条的时候，上个入读情况还是保留的
if(flag != 3) {
step = i+1;//记录当前是第几条关系
ok = true;
}
}//三种情况
if(flag == 2) {
printf("Inconsistency found after %d relations.
", step);
} else if(flag == 3){
printf("Sorted sequence cannot be determined.
");
} else {
ans[cnt] = ' ';
printf("Sorted sequence determined after %d relations: %s.
", step, ans);
}
}
}