【作业】Kitchen Plates（拓扑排序）

题目链接：https://vjudge.net/contest/345791#problem/O

【问题描述】

　　You are given 5 different sizes of kitchen plates. Each plate is marked with a letter A, B, C,D, or E. You are given 5 statements comparing two different plates, you need to rearrange the plates from smallest size to biggest size. For example: the sizes of these plates.

 输入：

The input consist of 5 lines. In each line there will be 3 characters, the first and last character will be either A, B, C, D, or E and the middle character will be either > or < describing the comparison between two plates sizes. No two plates will be equal.

输出：

The output consist of 55 characters, the sorted order of balls from smallest to biggest plate. Otherwise, if the statements are contradicting print impossibleimpossible. If there are multiple answers, print any of them.

样例输入;

D>B
A>D
E<C
A>B
B>C

B>E
A>B
E>A
C<B
D<B

样例输出：

ECBDA

impossible

试题分析：
　　给出5个大小关系，求其中一种满足从小到大的排列方式。因为输入并没有保证一定可以确定每两个Plate之间的大小关系，无法准确确定大小关系。题意只需要输出一种满足升序的排列即可，即用拓扑排序。
代码如下：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<string>
#include<map>
#define mem(a, b) memset(a, b, sizeof(a))
const int MAXN = 10;
using namespace std;

char s[MAXN];
int in[MAXN], out[MAXN];
int head[MAXN], cnt;
map<char, int> mp;
map<int, char> mpp;
string ans;

struct Edge
{
    int to, next;
}edge[MAXN];

void add(int a, int b)
{
    cnt ++;
    edge[cnt].to = b;
    edge[cnt].next = head[a];
    head[a] = cnt;
}

void topo_sort()
{
    queue<int> Q;
    for(int i = 1; i <= 5; i ++)
        if(!in[i])
            Q.push(i);
    while(!Q.empty())
    {
        int a = Q.front();
        Q.pop();
        ans += mpp[a];
        for(int i = head[a]; i != -1; i = edge[i].next)
        {
            int to = edge[i].to;
            in[to] --;
            if(!in[to])
            {
                Q.push(to);
            }
        }
    }
    if(ans.length() != 5)
        printf("impossible
");
    else
        cout << ans << endl;
}

int main()
{
    mp['A'] = 1, mp['B'] = 2, mp['C'] = 3, mp['D'] = 4, mp['E'] = 5;
    mpp[1] = 'A', mpp[2] = 'B', mpp[3] = 'C', mpp[4] = 'D', mpp[5] = 'E';
    
    mem(head, -1), cnt = 0;
    ans = "";
    for(int i = 1; i <= 5; i ++)
    {
        scanf("%s", s);
        int x = mp[s[0]], y = mp[s[2]];
        if(s[1] == '<')  //A < B则连边 表示A比B小 
        {
            add(x, y);
            out[x] ++;
            in[y] ++;
        }
        else
        {
            add(y, x);
            out[y] ++;
            in[x] ++;
        }    
    }
    topo_sort();
    return 0;
}