POJ 1094 Sorting It All Out

POJ 1094 Sorting It All Out

[★★☆☆☆]拓扑排序

• 题目大意：

  该题题意明确，就是给定一组字母的大小关系判断他们是否能组成唯一的拓扑序列。是典型的拓扑排序，但输出格式上确有三种形式：

  1.该字母序列有序，并依次输出；

  2.该序列不能判断是否有序；

  3.该序列字母次序之间有矛盾，即有环存在。

  而这三种形式的判断是有顺序的：先判断是否有环（3），再判断是否有序（1），最后才能判断是否能得出结果（2）。注意：对于（2）必须遍历完整个图，而（1）和（3）一旦得出结果，对后面的输入就不用做处理了。

• 代码

#include 
#include 
#include 

using namespace std;

int n, m;

int gx[30][30];
int tgx[30][30];

bool u[30];

int que[30];
int ctq;

void cpg() {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            tgx[i][j] = gx[i][j];
        }
    }
}

int rd(int l){
    int res = 0;
    for (int i = 1; i <= n; i++) {
        res += tgx[i][l];
    }
    return res;
}
int cd(int l){
    int res = 0;
    for (int i = 1; i <= n; i++) {
        res += tgx[l][i];
    }
    return res;
}

int yh() {
    cpg();
    ctq = 0;
    int v = -1;
    for (int i = 1; i <= n; i++) {
        u[i] = 0;
    }

    while (1) {
        v = -1;
        bool up = 0;
        for (int i = 1; i <= n; i++) {
            if (u[i] == 0 && rd(i) == 0) {
                v = i; up = 1;
                break;
            }
        }
        if (up == 0) break;
        que[ctq++] = v;
        u[v] = 1;
        for (int i = 1; i <= n; i++) {
            tgx[v][i] = 0;
        }
    }
    if (ctq == n) return 0;
    else return 1;

}

int TOP() {
    if (yh()) return -1;
    cpg();
    ctq = 0;
    int v = -1;
    for (int i = 1; i <= n; i++) {
        u[i] = 0;
    }
    while (1) {
        v = -1;
        bool up = 0;
        for (int i = 1; i <= n; i++) {
            if (u[i] == 0 && rd(i) == 0) {
                if (v == -1) {
                    v = i; up = 1;
                }
                else return 0;                  //没有固定顺序
            }
        }
        if (up == 0) break;
        que[ctq++] = v;
        u[v] = 1;
        for (int i = 1; i <= n; i++) {
            tgx[v][i] = 0;
        }
    }
    if (ctq == n) return 1;
    else return -1;
}

void solve() {
    string ts;
    for (int js = 0; js < m; js++) {
        cin >> ts;
        int t1, t2;
        t1 = ts[0] - 'A' + 1;
        t2 = ts[2] - 'A' + 1;
        gx[t1][t2] = 1;
        int r = TOP();
        if (r == 0) continue;
        if (r == -1) {
            cout << "Inconsistency found after " << js+1 << " relations." << endl;
            for (js++; jscin >> ts;
            return;
        }
        if (r == 1) {
            cout << "Sorted sequence determined after " << js+1 << " relations: ";
            for (int i = 0; i < ctq; i++) {
                char tc = que[i];
                tc = tc + 'A' - 1;
                cout << tc;
            }
            cout << '.' << endl;
            for (js++; jscin >> ts;
            return;
        }
    }
    int r = TOP();
    cout << "Sorted sequence cannot be determined." << endl;
    return;
}

int main() {


    while ((cin >> n >> m) && n) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                gx[i][j] = 0;
            }
            u[i] = 0;
        }
        ctq = 0;
        solve();
    }
    return 0;
}

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