poj1094（拓扑排序应用）

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题意：给m个大小关系，确定n个字母的大小关系，如果矛盾或者大小关系确定，那么就结束判断，否则继续判断。
解题思路：对于每次输入的信息，进行一次拓扑排序。

代码如下：

#include<iostream>#include<algorithm>#include<stdio.h>#include<queue>#include<math.h>#include<cstring>#include<string>const int N = 30;#define inf 0x3f3f3f3f#define pi acos(-1.0)#define eps 10e-6using namespace std;int d[N],dd[N];vector<int> a[N];int res[N];int n,m;int tuopu(){    queue<int> q;    int i,cnt = 0;    int flag = 0;    while(1)    {        for(i = 0; i < n; i++)            if(dd[i] == 0){                q.push(i);                dd[i] = -1;            }        if(q.size() > 1) flag = 1;        if(q.size() == 0) break;        int now = q.front();        q.pop();        res[cnt++] = now;        int len = a[now].size();        for(i = 0; i < len; i++)            dd[a[now][i]]--;    }    if(cnt < n) return 1;//矛盾    if(flag) return 2;    return 3;}int main(){    while(scanf("%d%d",&n,&m))    {        if(n == 0 && m == 0)  break;        memset(d,0,sizeof(d));        int i;        for(i = 0; i <= n; i++) a[i].clear();        int ans, k = 2;        for(i = 1; i <= m; i++)        {            char s[10];            scanf("%s",s);            if(k == 2)            {                int u = s[0] - 'A';                int v = s[2] - 'A';                d[v]++;                a[u].push_back(v);                for(int j = 0; j < n; j++)                    dd[j] = d[j];                k = tuopu();                if(k != 2)                    ans = i;            }        }        if(k == 1) printf("Inconsistency found after %d relations.\n",ans);        else if(k == 2) printf("Sorted sequence cannot be determined.\n");        else{            printf("Sorted sequence determined after %d relations: ",ans);            for(i = 0; i < n; i++)                printf("%c",(char)('A'+res[i]));            printf(".\n");        }    }    return 0;}

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