HDU 2612 简单bfs

来源：互联网发布：informatica linux 编辑：程序博客网时间：2024/06/06 06:31

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn = 2E2 + 10;const int INF = 0x3f3f3f3f;const int dir[][2] = {1, 0, 0, 1, -1, 0, 0, -1};int n, m, mp[maxn][maxn][2], dis[maxn][maxn][2];char str[maxn][maxn];struct Node{int x, y, step;bool in(int flag){return (x >= 0 && x < n && y >= 0 && y < m && str[x][y] != '#' && mp[x][y][flag] == 0);}};void bfs(int x, int y, int flag){queue<Node>Q;Node Cur, Next;Cur.x = x; Cur.y = y; Cur.step = 0;mp[x][y][flag] = 1;Q.push(Cur);while (!Q.empty()){Cur = Q.front(); Q.pop();Next.step = Cur.step + 1;for (int i = 0; i < 4; i++){Next.x = Cur.x + dir[i][0];Next.y = Cur.y + dir[i][1];if (Next.in(flag)){mp[Next.x][Next.y][flag] = 1;if (str[Next.x][Next.y] == '@')dis[Next.x][Next.y][flag] = Next.step;Q.push(Next);}}}}int main(int argc, char const *argv[]){while (~scanf("%d%d", &n, &m) && m + n){int ans = INF;memset(dis, INF, sizeof(dis));memset(mp, 0, sizeof(mp));for (int i = 0; i < n; i++)scanf("%s", str[i]);for (int i = 0; i < n; i++)for (int j = 0; j < m; j++){if (str[i][j] == 'Y')bfs(i, j, 0);else if (str[i][j] == 'M')bfs(i, j, 1);}for (int i = 0; i < n; i++)for (int j = 0; j < m; j++)if (str[i][j] == '@')ans = min(dis[i][j][0] + dis[i][j][1], ans);printf("%d\n", ans * 11);}return 0;}

两个源点，多个汇点的最短路，用bfs就可以，虽然我觉得这个题目应该放在图论里。。。

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