HDU 2612 Find a way（bfs）

解析：

求2个点到KFC的距离之和，使其最小，可用2次BFS，分别求出2个点到各个KFC的最短距离，然后找出和最小的即可。

AC代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 205;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = { 0, 1, 0,-1};

char grid[N][N];
int dp[N][N][2];
bool vis[N][N][2];

struct Node {
    int x, y, step;
    Node() {}
    Node(int _x, int _y, int _step) {
        x = _x, y = _y, step = _step;
    }
}poi[2];

int n, m;
bool judge(int x, int y) {
    if(x < 0 || x >= n || y < 0 || y >= m || grid[x][y] == '#')
        return true;
    return false;
}

void bfs(int sx, int sy, int ch) {
    queue<Node> que;
    dp[sx][sy][ch] = 0;
    vis[sx][sy][ch] = true;
    que.push(Node(sx, sy, 0));
    int x, y, step;
    while(!que.empty()) {
        Node front = que.front();
        que.pop();
        for(int i = 0; i < 4; i++) {
            x = front.x + dx[i];
            y = front.y + dy[i];
            step = front.step + 1;
            if(judge(x, y) || vis[x][y][ch]) continue;
            vis[x][y][ch] = true;
            que.push(Node(x, y, step));
            if(grid[x][y] == '@') {
                dp[x][y][ch] = step;
            }
        }
    }
}

int main() {
    while(scanf("%d%d", &n, &m) != EOF) {
        memset(dp, 0, sizeof(dp));
        memset(vis, false, sizeof(vis));
        for(int i = 0; i < n; i++) {
            scanf("%s", grid[i]);
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == 'Y')
                    poi[0].x = i, poi[0].y = j;
                else if(grid[i][j] == 'M')
                    poi[1].x = i, poi[1].y = j;
            }
        }
        for(int i = 0; i <= 1; i++) {
            bfs(poi[i].x, poi[i].y, i);
        }
        int ans = INF;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(grid[i][j] == '@' && vis[i][j][0] && vis[i][j][1])
                    ans = min(ans, dp[i][j][0] + dp[i][j][1]);
            }
        }
        printf("%d\n", ans*11);
    }
    return 0;
}