Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions:39731 | Accepted: 13975 |
Descriptionnode
An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.
Inputios
Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.
Outputc++
For each problem instance, output consists of one line. This line should be one of the following three:
Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.
where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence.
Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.
where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence.
Sample Input閉包
4 6 A<B A<C B<C C<D B<D A<B 3 2 A<B B<A 26 1 A<Z 0 0
Sample Outputless
Sorted sequence determined after 4 relations: ABCD. Inconsistency found after 2 relations. Sorted sequence cannot be determined.
Sourceide
題意：this
給定關於n個字母的對應優先級關係，問是否能夠根據優先級對他們進行排序。而且要求輸出是在第幾組能夠得出結果。url
會有優先級不一致和沒法排序的狀況。spa
思路：.net
感受這題數據很迷啊，m的範圍都不給我都很差估計複雜度。
以及，要注意輸出時候的句號。特別是能夠sort的時候。
是一道傳遞閉包的問題，用鄰接矩陣建圖，若是$A<B$，則表示$A$到$B$有一條有向邊，$g['A']['B'] = 1$
每次添加一個關係，使用一次floyd計算可否得出結果，或者是否出現不一致。
輸出排序結果的時候，其實只須要比較每一個點的入度，按照入度從大到小排序就好了。
由於最後的矩陣是一個傳遞閉包，最小的那個字母確定小於其餘全部字母，也就是說他這一行會有$n-1$個$1$
1 #include<iostream> 2 //#include<bits/stdc++.h> 3 #include<cstdio> 4 #include<cmath> 5 #include<cstdlib> 6 #include<cstring> 7 #include<algorithm> 8 #include<queue> 9 #include<vector> 10 #include<set> 11 #include<climits> 12 using namespace std; 13 typedef long long LL; 14 #define N 100010 15 #define pi 3.1415926535 16 #define inf 0x3f3f3f3f 17 18 int n, m; 19 int g[30][30], tmpg[30][30]; 20 21 int floyd() 22 { 23 for(int i = 0; i < n; i++){ 24 for(int j = 0; j < n; j++){ 25 tmpg[i][j] = g[i][j]; 26 } 27 } 28 for(int k = 0; k < n; k++){ 29 for(int i = 0; i < n; i++){ 30 for(int j = 0; j < n; j++){ 31 tmpg[i][j] |= tmpg[i][k] & tmpg[k][j]; 32 if(tmpg[i][j] == tmpg[j][i] && tmpg[i][j] == 1 && i != j){ 33 return -1; 34 } 35 } 36 } 37 } 38 for(int i = 0; i < n; i++){ 39 for(int j = 0; j < n; j++){ 40 if(tmpg[i][j] == tmpg[j][i] && tmpg[i][j] == 0 && i != j){ 41 return 0; 42 } 43 } 44 } 45 return 1; 46 } 47 48 struct node{ 49 int deg; 50 char ch; 51 }; 52 bool cmp(node a, node b) 53 { 54 return a.deg > b.deg; 55 } 56 57 void print() 58 { 59 //int deg[30]; 60 //memset(deg, 0, sizeof(deg)); 61 node character[30]; 62 for(int i = 0; i < n; i++){ 63 character[i].deg = 0; 64 character[i].ch = 'A' + i; 65 } 66 for(int i = 0; i < n; i++){ 67 for(int j = 0; j < n; j++){ 68 if(tmpg[i][j]){ 69 character[i].deg++; 70 } 71 } 72 } 73 sort(character, character + n, cmp); 74 for(int i = 0; i < n; i++){ 75 printf("%c", character[i].ch); 76 } 77 printf(".\n"); 78 79 /*queue<int>que; 80 for(int i = 0; i < n; i++){ 81 printf("%d\n", deg[i]); 82 if(!deg[i]){ 83 que.push(i); 84 break; 85 } 86 } 87 for(int i = 0; i < n; i++){ 88 int x = que.front();que.pop(); 89 printf("%c", x + 'A'); 90 for(int k = 0; k < n; k++){ 91 if(tmpg[k][x])deg[k]--; 92 if(!deg[k])que.push(k); 93 } 94 }*/ 95 } 96 97 int main() 98 { 99 while(scanf("%d%d", &n, &m) != EOF && n || m){ 100 memset(g, 0, sizeof(g)); 101 /*for(int i = 0; i < n; i++){ 102 g[i][i] = 1; 103 }*/ 104 105 int i; 106 bool dont = false; 107 for(i = 1; i <= m; i++){ 108 char a, b; 109 getchar(); 110 scanf("%c<%c", &a, &b); 111 g[a - 'A'][b - 'A'] = 1; 112 if(!dont){ 113 int flag = floyd(); 114 if(flag == -1){ 115 printf("Inconsistency found after %d relations.\n", i); 116 dont = true; 117 } 118 else if(flag == 1){ 119 printf("Sorted sequence determined after %d relations: ", i); 120 print(); 121 dont = true; 122 } 123 } 124 } 125 if(i > m && !dont){ 126 printf("Sorted sequence cannot be determined.\n"); 127 } 128 } 129 return 0; 130 }