poj1024:Tester Program给出最短路径求对应迷宫

题意大概如下：給出最短路径，要求一个迷宫，使得该迷宫的唯一最短路径即为此路径，且迷宫中的每一堵墙都不能多余。然后给出一种方案，要你判断此方案是否正确。

我的算法是：先bfs得出一条最短路径和最短路径的总数，若大于1或者跟输入的最短路径不同均不可以。然后计算对于每一个墙，如果可以通过该墙的话最短路径是多少。

如果有算出大于输入的最短路径的(也即是该墙多余)，也不可以。算法完毕；

AC代码如下：

# include <stdlib.h># include <string.h># include <stdio.h># define DEBUGint w, h;int used[110][110];int wall[100][100][4];int unuse[100][100];int from[100][100];char way[10000];int data[100][100];int m;int ex, ey;int xx[] = {1,-1,0,0};int yy[] = {0,0,1,-1};int is_block(int x1, int y1, int x2, int y2) {if (x1 >= w || x1 < 0) return 1;if (x2 >= w || x2 < 0) return 1;if (y1 >= h || y1 < 0) return 1;if (y2 >= h || y2 < 0) return 1;for (int i = 0; i < 4; ++ i) {if (wall[x1][y1][i]) {int x = wall[x1][y1][i] / 100;int y = wall[x1][y1][i] % 100;if (x == x2 && y == y2) return 1;}if (wall[x2][y2][i]) {int x = wall[x2][y2][i] / 100;int y = wall[x2][y2][i] % 100;if (x == x1 && y == y1) return 1;}}return 0;}int make_the_way() {int k = (int)strlen(way);char away[k + 1];memset(away, 0, sizeof(away));int xxx = ex; int yyy = ey;for (int i = k - 1; i >= 0; -- i) {if (xxx - from[xxx][yyy] / 100 == -1) {away[i] = 'L';if (way[i] != 'L') return -1;}if (xxx - from[xxx][yyy] / 100 == 1) {away[i] = 'R';if (way[i] != 'R') return -1;}if (yyy - from[xxx][yyy] % 100 == -1) {away[i] = 'D';if (way[i] != 'D') return -1;}if (yyy - from[xxx][yyy] % 100 == 1) {away[i] = 'U';if (way[i] != 'U') return -1;}int t = from[xxx][yyy];xxx = t / 100;yyy = t % 100;data[xxx][yyy] = k - i;}away[k] = '\0';return 0;}int bfs(int, int, int, int);int abs(int a) {return a > 0 ? a : -a;}int min(int a, int b) {return a > b ? b : a;}int is_unuse() {int k = strlen(way);for (int x1 = 0; x1 < w; ++ x1) {for (int y1 = 0; y1 < h; ++ y1) {for (int x2 = 0; x2 < w; ++ x2) {for (int y2 = 0; y2 < h; ++ y2) {if (is_block(x1, y1, x2, y2)) {int b1 = abs(bfs(0, 0, x1, y1));int b2 = abs(bfs(0, 0, x2, y2));int e1 = abs(bfs(x1, y1, ex, ey));int e2 = abs(bfs(x2, y2, ex, ey));int t = min(b1 + e2 + 1, b2 + e1 + 1);if (t > k) return 1;}}}}}return 0;}int bfs(int bx, int by, int ex, int ey) {if(bx == ex && by == ey) return 0;int head = 0;int tail = 0;int step = 0;int queue[10010];int ans = 0;memset(queue, 0, sizeof(queue));memset(used, 0, sizeof(used));queue[0] = bx * 100 + by;while (tail >= head) {++ step;int size = tail - head;for (int i = head; i <= size + head; ++ i) {int x1, y1, x2, y2;x1 = queue[i] / 100;y1 = queue[i] % 100;used[x1][y1] = 1;for (int j = 0; j < 4; ++ j) {x2 = x1 + xx[j];y2 = y1 + yy[j];if (!used[x2][y2] && !is_block(x1, y1, x2, y2)) {from[x2][y2] = queue[i];//printf("%d is from %d\n", x2 * 100 + y2, from[x2][y2]);used[x2][y2] = 1;++ tail;queue[tail] = x2 * 100 + y2;if (x2 == ex && y2 == ey) {++ ans;}}}}if (ans > 1) return -step;if (ans == 1){ return step;}head += size + 1;}return 10010;}int main() {int ts;for (scanf("%d", &ts); ts; -- ts) {scanf("%d %d\n%s\n%d", &w, &h, way, &m);memset(from, 0, sizeof(from));memset(unuse, 0, sizeof(unuse));memset(wall, 0, sizeof(wall));for (int i = m; i; -- i) {int x1, y1, x2, y2;scanf("%d %d %d %d", &x1, &y1, &x2, &y2);for (int i = 0; i < 4; ++ i) {if(!wall[x1][y1][i]) {wall[x1][y1][i] = x2 * 100 + y2;break;}}unuse[x1][y1] = unuse[x2][y2] = 1;}ex = 0; ey = 0;for (int i = 0; i < (int)strlen(way); ++ i) {switch(way[i]) {case 'R' : ++ ex; break;case 'U' : ++ ey; break;case 'L' : -- ex; break;case 'D' : -- ey; break;}}int t = bfs(0, 0, ex, ey);if (!make_the_way() && !is_unuse()){printf("CORRECT\n");}else {printf("INCORRECT\n");}}return 0;}