USACO2.4.1 The Tamworth Two (ttwo)

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如果初始时两人四周都有墙则直接输出0。
然后开始模拟。如果两人都回到了出发点则输出0。
还有就是如果两人都走了很多步就认为无法相遇输出0。

否则如果途中相遇了就输出。

/*ID:shijiey1PROG:ttwoLANG:C++*/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct Point{int x, y;};int arr[12][12];Point john;Point cow;Point posj;Point posc;Point sj;Point sc;int main() {freopen("ttwo.in", "r", stdin);freopen("ttwo.out", "w", stdout);char c;for (int i = 0; i <= 11; i++) {arr[0][i] = arr[i][0] = 1;arr[11][i] = arr[i][11] = 1;}for (int i = 1; i <= 10; i++) {for (int j = 1; j <= 10; j++) {scanf("%c", &c);if (c == '*') arr[i][j] = 1;if (c == 'F') {john.x = sj.x = i;john.y = sj.y = j;}if (c == 'C') {cow.x = sc.x = i;cow.y = sc.y = j;}}getchar();}posj.x = posc.x = -1;posj.y = posc.y = 0;if (arr[john.x + 1][john.y] && arr[john.x - 1][john.y] && arr[john.x][john.y + 1] && arr[john.x][john.y - 1]|| arr[cow.x + 1][cow.y] && arr[cow.x - 1][cow.y] && arr[cow.x][cow.y + 1] && arr[cow.x][cow.y - 1]) {printf("0\n");return 0;}int cnt = 0;while (true) {// printf("%d %d | %d %d\n", john.x, john.y, cow.x, cow.y);if (cnt > 2000000) {printf("0\n");return 0;}if (arr[john.x + posj.x][john.y + posj.y] == 1) {int t = posj.y;if (!posj.x) posj.y = 0;else posj.y = posj.y - posj.x;posj.x = t;} else {john.x += posj.x;john.y += posj.y;}if (arr[cow.x + posc.x][cow.y + posc.y] == 1) {int t = posc.y;if (!posc.x) posc.y = 0;else posc.y = posc.y - posc.x;posc.x = t;} else {cow.x += posc.x;cow.y += posc.y;}if (john.x == cow.x && john.y == cow.y) {printf("%d\n", cnt + 1);return 0;}cnt++;}return 0;}

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