Breadth-first search-minimum number of connections and breadth-first connections

Breadth-first search-minimum number of connections and breadth-first connections

When you travel to Hainan with your family, but your city has not directly arrived in Hainan, but you have collected a lot of flight information, and now you want to find a way to ride, minimize the number of connections

How can this problem be solved?

Assume that your city is in City 1 and Hainan is in city 5. The relationship is as follows:

How can we obtain the minimum number of transfers from City 1 to city 5? At this time, the content of this explanation is used,Search for breadth first!

First, we should useAdjacent matrixOrTwo-dimensional arrayTo access the relationship between vertices.

The extended search requires a queue to store the relationships of each extension.

First of all, we will join City 1. Through city 1, we can expand City 2 and City 3, and city 2 can expand City 3 and City 4. As City 3 is already in the queue, you only need to team City 4.

Next, city 3 can expand to City 4 and city 5. Because City 4 is already in the queue, you only need to team city 5. Now that city 5 has been found, the algorithm ends.

The Code is as follows:

# Include <stdio. h> struct node {int x; // city no. int s; // Number of connections} que [2501]; int main () {int e [51] [51] = {0}, book [51] = {0}; // store the relationship between the graph and the flag array int head, tail; int I, j, n, m, a, B, cur, start, end, flag = 0; scanf ("% d", & n, & m, & start, & end); // n indicates the number of cities, m indicates the relationship, start indicates the start city, and end indicates the destination city // initializes the two-dimensional matrix for (I = 1; I <= n; I ++) {for (j = 1; j <= n; j ++) if (I = j) e [I] [j] = 0; // This is the distance from yourself to yourself. else e [I] [j] = 0x3f3f3f3f; // hexadecimal, represents an infinite constant} // read the flight between cities for (I = 1; I <= m; I ++) {sca Nf ("% d", & a, & B); e [a] [B] = 1; // The undirected graph is a bidirectional e [B] [a] = 1;} // initialize the queue head = 1; tail = 1; // starts from the city of start, add the start city to the queue que [tail]. x = start; que [tail]. s = 0; tail ++; book [start] = 1; // The city marked with start is already in the queue. // when the queue is not empty, loop while (head <tail) {cur = que [head]. x; // number of the first city in the queue for (j = 1; j <= n; j ++) // try from 1-N in turn {// determine whether the city j has flights from the city cur to the city j and whether the city j is in the queue if (e [cur] [j]! = 0x3f3f3f3f & book [j] = 0) {// if there is a flight from cur to city j and city j is not in the queue, then j enters the queue que [tail]. x = j; que [tail]. s = que [head]. s + 1; // Number of connections + 1 tail ++; book [j] = 1; // change the flag to prevent reuse} // stop scaling when the target city is reached, exit loop if (que [tail]. x = end) {flag = 1; break ;}}if (flag = 1) break; head ++; // after the extension ends, head ++ can continue expansion} printf ("% d \ n", que [tail-1]. s); // Because tail points to the next position at the end of the queue, 1 return 0;}/* 5 7 1 51 21 32 32 43 43 54 5 */