POJ - 1094 Sorting It All Out 【拓扑排序】

题目链接

http://poj.org/problem?id=1094

题意

给出n个点，m对关系

判断 是否能够有一个确定的排列，或者矛盾，或者没有确定的排列

思路

在代码下面的注释中

AC代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define CLR(a, b) memset(a, (b), sizeof(a));
#define pb push_back
#define bug puts("***bug***");
#define X first
#define Y second
#define L(on) (on<<1)
#define R(on) (L(on) | 1)
#define all(x) x.begin(), x.end()
#define stack_expand #pragma comment(linker, "/STACK:102400000,102400000");
#define syn_close   ios::sync_with_stdio(false);cin.tie(0);
//#define bug 
//#define gets gets_s 

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair  pll;
typedef pair <string, int> psi;
typedef pair <string, string> pss;
typedef pair <double, int> pdi;

const double PI = acos(-1.0);
const double EI = exp(1.0);
const double eps = 1e-8;

const int INF = 0x3f3f3f3f;
const int maxn = 30 + 10;
const int MOD = 1e9 + 7;

int G[maxn][maxn];

int du[maxn];

int n, m;

void Flyod()
{
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                G[i][j] = (G[i][j] || (G[i][k] && G[k][j]));
}

bool judge()
{
    for (int i = 1; i <= n; i++)
        if (G[i][i] == 1)
            return false;
    return true;
}

vector <int> ans;

vector <int> g[maxn];

void toposort()
{
    ans.clear();
    queue <int> q;
    for (int i = 1; i <= n; i++)
        if (du[i] == 0)
            q.push(i);
    while (!q.empty())
    {
        int id = q.front(); q.pop(); ans.pb(id);
        int len = g[id].size();
        for (int i = 0; i < len; i++)
            if ((--du[g[id][i]]) == 0)
                q.push(g[id][i]);
    }
}

void print()
{
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            printf("%d%c", G[i][j], j == n ? '\n' : ' ');
}

void clear()
{
    for (int i = 1; i <= n; i++)
    {
        du[i] = G[i][i] = 0;
        g[i].clear();
        for (int j = i + 1; j <= n; j++) G[i][j] = G[j][i] = 0;
    }
}

int main()
{
    while (~scanf("%d%d", &n, &m) && (n || m))
    {
        clear();
        int x, y;
        int flag = 0;
        int vis = 0;
        char s[5];
        for (int i = 0; i < m; i++)
        {
            scanf(" %s", s);
            x = s[0] - 'A' + 1;
            y = s[2] - 'A' + 1;
            if (vis == 0)
            {
                g[x].pb(y);
                du[y]++;
            }
            G[x][y] = 1;
            Flyod();
            if (judge() == false && flag == 0 && vis == 0)
                flag = i + 1;
            else if (vis == 0)
            {
                int tot = 0;
                for (int j = 1; j <= n; j++)
                {
                    int res = 0;
                    for (int k = 1; k <= n; k++)
                        res += G[j][k] || G[k][j];
                    if (res == n - 1)
                        tot++;
                }
                if (tot == n)
                    vis = i + 1;
            }
        }
        if (flag != 0)
        {
            printf("Inconsistency found after %d relations.\n", flag);
            continue;
        }
        if (vis == 0)
        {
            printf("Sorted sequence cannot be determined.\n");
            continue;
        }
        toposort();
        printf("Sorted sequence determined after %d relations: ", vis);
        for (int i = 0; i < n; i++)
            printf("%c", (ans[i] - 1 + 'A'));
        printf(".\n");
    }
}

/*
    如果 在当前给出的数据以上（包含当前这条数据） 
    已经能够确定答案 那么下面的数据是不用处理的  即使下面的数据会对答案造成影响

    0.如何确定关系矛盾
    我们用二分图 来构造一个有向图 G[i][j] 表示 i < j
    然后用弗洛伊德跑一下

    显然 如果给出的关系矛盾 一定是下面两种情况 
        0. A