poj1094Sorting It All Out

刚遇到这道题时我就想着把它建立成一个图，然后每输入一条边就就深度遍历一下。

判断是否有环和是否能判断出大小顺序。。。

但是之后才知道一般递归都很耗时间。。。

尽管我加入了几个强大的剪枝。而且在discuss里面的数据都通过了。

不过还是超时。。。。

我特别注明一下，下面的代码是超时的代码额！！

希望大神能帮我修改一下ac。。小弟感激不尽！！！

#include<stdio.h>#include<string.h>#include<stdlib.h>int map[30][30];int chvist[30],order[30],isnothead[30],rudu[30];int n,m,vist[30],num;int DFS(int j,int layer,int xx[]){//vist[j] = 1;xx[layer] = j;if(layer == n-1){for(int j=0;j<n;j++)order[j] = xx[j];return 1;}for(int i=0;i<n;i++)if(map[j][i]){if(!vist[i]){vist[i] = 1;int ab = DFS(i,layer+1,xx);vist[i] = 0;if(ab != 0)return ab;}elsereturn -1;}return 0;}int main(){char ch[5];int xx[30];while(scanf("%d%d",&n,&m) && n && m){int isnotco = 1;int ring_or_order,times;memset(map,0,sizeof(map));memset(chvist,0,sizeof(chvist));memset(isnothead,0,sizeof(isnothead));memset(rudu,0,sizeof(rudu));for(int i=0;i<m;i++){scanf("%s",ch);chvist[ch[0]-'A'] = 1;chvist[ch[2]-'A'] = 1;rudu[ch[2]-'A']++;if(ch[1] == '<'){map[ch[0]-'A'][ch[2]-'A'] = 1;isnothead[ch[0]-'A'] = 1;isnothead[ch[2]-'A'] = 0;}else{map[ch[1]-'A'][ch[0]-'A'] = 1;isnothead[ch[0]-'A'] = 0;isnothead[ch[2]-'A'] = 1;}int rudunum = 0;for(int tt=0;tt<n;tt++)if(rudu[tt]==0)rudunum++;for(int j=0;j<n;j++)if(chvist[j] && isnotco && isnothead[j]==1 && rudunum==1){num = 0;memset(vist,0,sizeof(vist));vist[j] = 1;         int flag = DFS(j,0,xx); if(flag == -1) { isnotco = 0; ring_or_order = 1; times = i+1; } else if(flag == 1) { isnotco = 0; ring_or_order = 2; times = i+1; }}}if(isnotco)printf("Sorted sequence cannot be determined.\n");else{if(ring_or_order==1)printf("Inconsistency found after %d relations.\n",times);else{printf("Sorted sequence determined after %d relations: ",times);for(int s=0;s<n;s++)printf("%c",order[s]+'A');printf(".\n");}}}return 0;}

下面是用拓扑排序0ms  ac的链接。。前辈的

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