CF 广搜（基础）

1 2
.#...
.#.#.
...#.

output

15

input

3 3
4 5
..#
.#.
#..

output

IMPOSSIBLE

http://codeforces.com/gym/100796/problem/D

题目大意：

刚开始是在小镇里面，小镇到荒地要a元，荒地到小镇要b元。问，怎么走才能花费的最少

思路：

基础的广搜。。。不过我要吐槽。。。为啥我第一种方法是wa了，害我重新又写了好一会儿TAT

#include
#include
#include
#include

using namespace std;

typedef long long ll;
typedef pair P;
const int maxn = 500 + 5;
int n, m;
int a, b;
char atlas[maxn][maxn];
int step[maxn][maxn];
int dx[4] = {1, 0, 0, -1};
int dy[4] = {0, 1, -1, 0};
int val[maxn][maxn];

void init(){
    memset(step, 0, sizeof(step));
    memset(atlas, 0, sizeof(atlas));
    memset(val, -1, sizeof(val));
    memset(step, -1, sizeof(step));
    scanf("%d%d", &a, &b);
    for (int i = 0; i < m; i++){
        scanf("%s", atlas[i]);
    }
    val[0][0] = a;
    queue 
 que;
    que.push(make_pair(0, 0));
    while (!que.empty()){
        P tmp = que.front();
        que.pop();
        for (int i = 0; i <= 3; i++){
            int x = tmp.first + dx[i];
            int y = tmp.second + dy[i];
            if (x < 0 || y < 0 || x >= m || y >= n) continue;
            if (atlas[x][y] == '#' || atlas[x][y] == '\0') continue;
            if (val[x][y] != -1) continue;
            if (val[tmp.first][tmp.second] == a){
                val[x][y] = b;
            }
            else val[x][y] = a;
            que.push(make_pair(x, y));
        }
    }
    /*
    printf("check\n");
    for (int i = 0; i < m; i++){
        for (int j = 0; j < n; j++){
            printf("%d ", val[i][j]);
        }
        printf("\n");
    }
    */
}

ll bfs(){
    queue 
 que;
    que.push(make_pair(0, 0));
    step[0][0] = 0;
    while (!que.empty()){
        P tmp = que.front();
        que.pop();
        for (int i = 0; i < 4; i++){
            int x = tmp.first + dx[i];
            int y = tmp.second + dy[i];
            if (x < 0 || y < 0 || x >= m || y >= n) continue;
            if (atlas[x][y] == '#' || atlas[x][y] == '\0') continue;
            if (step[x][y] != -1) continue;
            step[x][y] = step[tmp.first][tmp.second] + val[x][y];
            //printf("step[%d][%d] = %d\n", x, y, step[x][y]);
            que.push(make_pair(x, y));
        }
    }
    if (step[m-1][n-1] == -1) printf("IMPOSSIBLE\n");
    else printf("%d\n", step[m-1][n-1]);
}

int main(){
    while (scanf("%d%d", &n, &m) == 2){
        init();
        bfs();
    }
    return 0;
}