1199 problem B: size relationship

Floyd Warshall algorithm for finite set transitive closure (matrix implementation)

In fact, the triple loop.
Zzuoj 1199 Questions Link http://acm.zzu.edu.cn:8000/problem.php?id=1199
Problem B: Size relationshipTime limit:2 Sec Memory limit:128 MB
submit:148 solved:31
[submit][status][Web Board] Description

When we know a set of size relationships, we can tell if all relationships can be established, that is, there is no contradiction between relationships.

For example: A<b, A<c, b<c through this group of relations we can get a<b<c, all relations are established, there is no contradiction.

If a<b, B<c, c<a through the first two relations we get a<b<c, this relationship and c<a contradiction, all relations can not be set up at the same time.

Now that we know the M relationship, please determine whether the M relationship can be set up, set the output "YES", otherwise output "NO".

Input

Multiple sets of data, each set of data is as follows:

The first line has a number m. M represents the M-group relationship (1<=M<=400), and the next M-line has a relationship, denoted by two different letters and a symbol. (Enter a guaranteed letter between ' A '-' Z ', the relationship symbol is only >, <)

Output

Output "YES" or "NO" for each set of data.

Sample Input3a<ba<cb<c3a<bb<cC<aSample Output

YES

NO

/********************************************************* *************************************************** ***********/

Greater than and less than is actually a relationship, that is, the relationship between two elements, B>a can be described by a < B.

The question given is the relationship between the 26 letters, we can use 0 to 25 of the integer corresponding to A to Z, a two-dimensional array to express the relationship between elements, such as Re[0][25]==1 is 0 < 25, that is, a < Z. All that remains is the process of reasoning.

People who have studied discrete mathematics (Shanghai Science and Technology literature) know that there is a property called transitivity in the nature of the relationship in set theory, and that the relationship between < is exactly the same, and of course, less than satisfying reflexivity and symmetry is the key to our contradiction.

In the closure operation of the relationship, there is a closure operation called the transitive closure r+ of the relationship R, and this r+ is passed, that is to say, if a < b and B < C are found in r+, a < C will be found, that is to say r+ is reasoning After the relationship is obtained.

The algorithm for finding r+:

for (i=0; i<26;++i) {
for (J=0;J<26;++J) {
if (re[j] [i]) {
for (K=0;k<26;++k) {
Re[j] [K] +=re[I [K];
}
}
}
}

That's all, thanks Warshall.

Incidentally, this Floyd Warshall and graph theory inside of the classic Floyd algorithm is a person, in fact, this problem can also be completely in the form of graph theory to do

/***************************************/

AC Code:

# include <iostream>
# include <string.h>
# include <stdlib.h>
# include <math.h>
# include <stdio.h>
# include <algorithm>
# include <stack>

int m,re[27 [27];
Char a,b,c,in[4];

int main () {
using namespace Std;
int i,j,k;
while (~SCANF ("%d", &m)) {

memset (re,0,sizeof (re));

for (i=0; i<m; ++i) {
scanf ("%s", in);
a=in[0];
c=in[1];
b=in[2];

if (c== ' < ') {
re[A-' a ' [B ' a ']=1;
}else if (c== ' > ') {
re[b ' A ' [a ' a ']=1;
}
}



for (i=0; i<26;++i) {
for (J=0;J&LT;26;++J) {
if (re[j] [i]) {
for (K=0;k<26;++k) {
Re[j] [k]+=re[i] [K];
}
}
}
}

for (I=0;i<26;++i) {
for (J=0;J&LT;26;++J)
if (re[I [J]&&re[J] [i]) {
cout << "no\n";
Goto KKK;
}
}

cout << "yes\n";
KKK:;
}

return 0;
}

Reward

1199 problem B: size relationship