314 - Robot(用Dijkstra怎么还ac不了)

最新推荐文章于 2024-08-07 17:26:28 发布

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本文探讨了一种使用优先级队列和动态规划算法解决二维迷宫问题的方法，通过设置起始点和终点坐标，利用网格结构和方向数组来追踪最短路径。算法在遍历过程中避免重复计算，显著提高了求解效率。

#include <cstdio>
#include <cstring>
#include <queue>
#include <string>

using namespace std;

const int N = 55;
const int M = 4;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
const int INF = 0x3f3f3f3f;

struct HeapNode
{
    int x, y, d, dir;
    bool operator < (const HeapNode &other) const
    {
        return d > other.d;
    }
};

int n, m;
bool vis[M][N][N];
int grid[N][N];
int curdir;
int srcx, srcy, dstx, dsty;
char buf[N];
int d[M][N][N];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("e:\\uva_in.txt", "r", stdin);
#endif

    while (scanf("%d%d", &m, &n) == 2 && (n || m)) {
        memset(grid, 0x00, sizeof(grid));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                scanf("%d", &grid[i][j]);
            }
        }

        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (grid[i][j]) {
                    grid[i + 1][j] = grid[i][j + 1] = grid[i + 1][j + 1] = 1;
                }
            }
        }
        scanf("%d%d%d%d%s", &srcx, &srcy, &dstx, &dsty, buf);
        string s = buf;
        if (s == "north") curdir = 0;
        else if (s == "east") curdir = 1;
        else if (s == "south") curdir = 2;
        else curdir = 3;

        //printf("srcx=%d, srcy=%d, dstx=%d, dsty=%d, curdir=%d\n", srcx, srcy, dstx, dsty, curdir);

        if (grid[srcx][srcy] || grid[dstx][dsty]) {
            printf("-1\n");
            continue;
        }

        if (srcx < 0 || srcx >= m || srcy < 0 || srcy >= n) {
            printf("-1\n");
            continue;
        }

        if (dstx < 0 || dstx >= m || dsty < 0 || dsty >= n) {
            printf("-1\n");
            continue;
        }

        priority_queue<HeapNode> q;
        HeapNode node;
        bool found = false;
        node.x = srcx, node.y = srcy, node.d = 0, node.dir = curdir;
        q.push(node);
        memset(d, 0x3f, sizeof(d));
        d[curdir][srcx][srcy] = 0;
        memset(vis, false, sizeof(vis));

        while (!q.empty()) {
            HeapNode tmp = q.top(); q.pop();
            int curx = tmp.x, cury = tmp.y;
            int curdir = tmp.dir;

            //printf("curx=%d, cury=%d, curdir=%d, d=%d\n", curx, cury, curdir, tmp.d);

            if (curx == dstx && cury == dsty) {
                found = true;
                printf("%d\n", tmp.d);
                break;
            }

            if (vis[curdir][curx][cury]) continue;
            vis[curdir][curx][cury] = true;

            int newDir = (curdir + 1) % 4;
            int x = curx, y = cury;
            if (d[curdir][curx][cury] + 1 < d[newDir][curx][cury]) {
                d[newDir][curx][cury] = d[curdir][curx][cury] + 1;
                node.x = x, node.y = y, node.dir = newDir, node.d = d[newDir][curx][cury];
                q.push(node);
            }

            newDir = (curdir + 3) % 4;
            x = curx, y = cury;
            if (d[curdir][curx][cury] + 1 < d[newDir][curx][cury]) {
                d[newDir][curx][cury] = d[curdir][curx][cury] + 1;
                node.x = x, node.y = y, node.dir = newDir, node.d = d[newDir][curx][cury];
                q.push(node);
            }

            newDir = curdir;
            for (int i = 1; i <= 3; i++) {
                x = curx + dx[curdir] * i;
                y = cury + dy[curdir] * i;
                if (x < 0 || x >= m || y < 0 || y >= n) break;
                if (grid[x][y]) break;

                if (d[curdir][curx][cury] + 1 < d[newDir][x][y]) {
                    d[newDir][x][y] = d[curdir][curx][cury] + 1;
                    node.x = x, node.y = y, node.dir = newDir, node.d = d[newDir][x][y];
                    q.push(node);
                }
            }


        }

        if (!found) {
            printf("-1\n");
        }
    }

    return 0;
}