CodeForces 328B (反向BFS)-优快云博客

本文详细介绍了如何通过BFS算法解决Codeforces竞赛中的329B问题，该问题涉及从起点到终点的最短路径寻找，并结合了二维数组的动态规划思想，通过优化搜索过程提高了效率。

题目链接：http://codeforces.com/problemset/problem/329/B

#include <cstdio>

const int inf = (int)1e9;

const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};

char s[1111][1111];
int x[1111111], y[1111111];
int dist[1111][1111];

int main()
{
    int r, c;
    scanf("%d %d", &r, &c);
    for (int i = 0; i < r; i++) scanf("%s", s[i]);
    for (int i = 0; i < r; i++)
        for (int j = 0; j < c; j++) {
            dist[i][j] = inf;
            if (s[i][j] == 'E') {
                x[1] = i;
                y[1] = j;
                dist[i][j] = 0;
            }
        }
    int b = 1, e = 1;
    while (b <= e) {
        for (int j = 0; j < 4; j++) {
            int xk = x[b] + dx[j];
            int yk = y[b] + dy[j];
            if (xk >= 0 && yk >= 0 && xk < r && yk < c)
                if (s[xk][yk] != 'T' && dist[xk][yk] == inf) {
                    e++;
                    x[e] = xk;
                    y[e] = yk;
                    dist[xk][yk] = dist[x[b]][y[b]] + 1;
                }
        }
        b++;
    }
    int thr = -1;
    for (int i = 0; i < r; i++)
        for (int j = 0; j < c; j++)
            if (s[i][j] == 'S') thr = dist[i][j];
    int ans = 0;
    for (int i = 0; i < r; i++)
        for (int j = 0; j < c; j++)
            if (s[i][j] >= '0' && s[i][j] <= '9')
                if (dist[i][j] <= thr) ans += s[i][j] - '0';
    printf("%d\n", ans);
    return 0;
}