POJ 3026 Borg Maze wide search (BFS) + Minimum Spanning Tree, pojborg

POJ 3026 Borg Maze wide search (BFS) + Minimum Spanning Tree, pojborg

Starting from S, capture each A and find the shortest length of the total path. People at S and A can be divided into two people, but only one person can go at a time.

The idea is to first use BFS to find the distance between each point, build a graph, and then apply the minimum spanning tree template.

One-time. However, I think it is a little troublesome to judge the number.

# Include <iostream> # include <cstdio> # include <cstring> # include <queue> using namespace std; const int inf = 1000000; const int maxn = 205; char mapp [maxn] [maxn]; bool visit [maxn] [maxn]; bool vis [maxn]; int dx [] = {-1, 0, 1, 0 }; // Four Directions: int dy [] = {0,-, 1}; int mat [maxn] [maxn], dis [maxn]; int cou, ans; struct Node {int x; int y; int s; // record number from int num; // vertex} next, nodes [maxn]; // nodes [] Save all vertices void bfs (Node node, int a) {me Mset (visit, 0, sizeof (visit); node. s = 0; mapp [node. y] [node. x] = ''; // After the access, change 's' or 'A' to''. You do not need to access the queue again next time <Node> q; visit [node. x] [node. y] = true; q. push (node); while (! Q. empty () {node = q. front (); if (mapp [node. y] [node. x] = 'A') {if (A = 0) {nodes [cou] = node; mat [a] [cou] = mat [cou] [a] = node. s; cou ++; // number of records} else {int B; for (int j = 0; j ++) if (nodes [j]. x = node. x & nodes [j]. y = node. y) {B = nodes [j]. num; break;} mat [a] [B] = mat [B] [a] = node. s; // write the distance to the diagonal matrix} q. pop (); for (int I = 0; I <4; I ++) {next = node; next. x = node. x + dx [I]; next. y = node. y + dy [I]; if (! Visit [next. x] [next. y] & (mapp [next. y] [next. x] = ''| mapp [next. y] [next. x] = 'A') {visit [next. x] [next. y] = true; next. s = node. s + 1; q. push (next) ;}}return ;}bool prim () {memset (vis, 0, sizeof (vis); for (int I = 0; I <cou; I ++) dis [I] = (I = 0? 0: inf); ans = 0; for (int I = 0; I <cou; I ++) {int temp = inf, k = 0; for (int j = 0; j <cou; j ++) {if (! Vis [j] & dis [j] <temp) {temp = dis [j]; k = j ;}} if (temp = inf) return false; vis [k] = true; ans + = temp; for (int j = 0; j <cou; j ++) {if (! Vis [j] & dis [j]> mat [k] [j]) dis [j] = mat [k] [j] ;}} return true ;} int main () {// freopen ("in.txt", "r", stdin); int n; scanf ("% d", & n); while (n --) {int x, y; memset (mapp, 0, sizeof (mapp); scanf ("% d", & x, & y ); char c [maxn] = {0}; gets (c); for (int I = 0; I <y; I ++) gets (mapp [I]); for (int I = 0; I <x; I ++) for (int j = 0; j <y; j ++) if (mapp [I] [j] ='s ') {nodes [0]. y = I; nodes [0]. x = j;} bool flag = true; cou = 1; for (int I = 0; I <cou; I ++) {bfs (nodes [I], I ); if (flag) {for (int j = 0; j <cou; j ++) nodes [j]. num = j; // number each vertex. Run only once flag = false ;}} prim (); printf ("% d \ n", ans );} return 0 ;}