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HDU - 2612 Find a way(BFS + 枚举)

2015-10-10 11:19 417 查看
题目大意:有两个人要去肯德基见面,问两个人到达任意一个肯德基的最短时间和

解题思路:先BFS找出每个人到达任意一个肯德基的最短时间,然后再判断即可,代码写搓了…

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using namespace std;
const int N = 210;
const int INF = 0x3f3f3f3f;

struct Node {
int x, y, time;
Node() {}
Node(int x, int y, int time): x(x), y(y), time(time) {}
};

int MinY

, MinM

;
bool visY

, visM

;
char map

;
int n, m, Yx, Yy, Mx, My;
int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};

void init() {
for (int i = 1; i <= n; i++) {
scanf("%s", map[i] + 1);
for (int j = 1; j <= m; j++)
if (map[i][j] == 'Y') {
Yx = i;
Yy = j;
}
else if (map[i][j] == 'M') {
Mx = i;
My = j;
}
}
}

void bfs1() {

memset(visM, 0, sizeof(visM));
queue<Node> q;
q.push(Node(Mx, My, 0));
visM[Mx][My] = 0;
MinM[Mx][My] = 0;
while (!q.empty()) {
int curx = q.front().x, cury = q.front().y, curt = q.front().time;
q.pop();
for (int i = 0; i < 4; i++) {
int xx = curx + dir[i][0];
int yy = cury + dir[i][1];
if (xx <= 0 || xx > n || yy <= 0 || yy > m || visM[xx][yy] || map[xx][yy] == '#') continue;
q.push(Node(xx, yy, curt + 1));
visM[xx][yy] = true;
MinM[xx][yy] = curt + 1;
}
}
}

void bfs2() {

memset(visY, 0, sizeof(visY));
queue<Node> q;
q.push(Node(Yx, Yy, 0));
visY[Yx][Yy] = 0;
MinY[Yx][Yy] = 0;
while (!q.empty()) {
int curx = q.front().x, cury = q.front().y, curt = q.front().time;
q.pop();
for (int i = 0; i < 4; i++) {
int xx = curx + dir[i][0];
int yy = cury + dir[i][1];
if (xx <= 0 || xx > n || yy <= 0 || yy > m || visY[xx][yy] || map[xx][yy] == '#') continue;
q.push(Node(xx, yy, curt + 1));
visY[xx][yy] = true;
MinY[xx][yy] = curt + 1;
}
}
}

void solve() {

bfs1();
bfs2();
int Min = INF;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
if (map[i][j] == '@') {
Min = min(MinY[i][j] + MinM[i][j], Min);
}
printf("%d\n", Min * 11);
}

int main() {
while (scanf("%d%d", &n, &m) != EOF) {
init();
solve();
}
return 0;
}
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