Day11 HDUOJ2612广度优先搜索求最小路径

Posted 2021-05-03 cc的心理活动

//hduoj 2612/*The input contains multiple test cases.Each test case include, first two integers n, m. (2<=n,m<=200).Next n lines, each line included m character.‘Y’ express yifenfei initial position.‘M’ express Merceki initial position.‘#’ forbid road;‘.’ Road.‘@’ KCF4 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#*/#include<iostream>#include<string>#include<queue>#define M 250using namespace std;int n,m;char a[M][M];bool visited[M][M];int dist[M][M];struct p { int row; int list; int dist;};
void BFS(queue<p> Q) { int kfcnum = 0; memset(visited, false, M * M * sizeof(bool)); int d[M][M];
 while (!Q.empty()) { p t = Q.front(); Q.pop();
 if (a[t.row][t.list] == '@') dist[t.row][t.list] += t.dist;
 if (t.list < m&&!visited[t.row][t.list+1]&&a[t.row][t.list+1]!='#') { p rt; rt.row = t.row; rt.list = t.list + 1; rt.dist = t.dist + 1; Q.push(rt); visited[t.row][t.list+1] = true; } if (t.row < n && !visited[t.row+1][t.list] && a[t.row+1][t.list] != '#') { p und; und.row = t.row+1; und.list = t.list; und.dist = t.dist + 1; Q.push(und); visited[t.row+1][t.list] = true; } if (t.list >0 && !visited[t.row][t.list - 1] && a[t.row][t.list - 1] != '#') { p lf; lf.row = t.row; lf.list = t.list - 1; lf.dist = t.dist + 1; Q.push(lf); visited[t.row][t.list-1] = true; } if (t.row>0 && !visited[t.row-1][t.list] && a[t.row-1][t.list] != '#') { p up; up.row = t.row-1; up.list = t.list; up.dist = t.dist + 1; Q.push(up); visited[t.row-1][t.list] = true;    }  }}
int main() { while (cin >> n) { memset(dist, 0, M * M * sizeof(int)); int numkfc = 0; queue<p> Qy; queue<p> Qm; cin >> m; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) cin >> a[i][j];
 for(int i=0;i<n;i++) for (int j = 0; j < m; j++) { if (a[i][j] == '@') numkfc++; if (a[i][j] == 'Y') { p y; y.dist = 0; y.row = i; y.list = j; Qy.push(y); } if (a[i][j] == 'M') { p ml; ml.dist = 0; ml.row = i; ml.list = j; Qm.push(ml); } } BFS(Qy); BFS(Qm); /*for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) cout << dist[i][j]; cout << endl; }*/ int min = 10000; for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) if (dist[i][j] != 0 && dist[i][j] < min) min=dist[i][j];        cout << min * 11 << endl; }}

利用广度优先搜索可以实现单源最短路径计算，类似填表的思想每次遍历都在距离上加一。