挺不錯的題目，很鍛煉代碼能力和調試能力~

題意：初始格子狀態固定，給你移動后格子的狀態，問最少需要多少步能到達，如果步數大于30，輸出-1。

由于單向搜索狀態太多，搜到二十幾就會爆了，所以應該想到雙向廣搜。

對于每一個格子，我只需要記錄上面和前面的格子顏色，因為格子左右移動前面顏色不變上面顏色變了，前后移動的話只是前面和上面的顏色交換了而已，不過寫的太挫了。。。應該直接進制壓縮一下。。。我還用了兩個數來表示前面和上面的顏色，下次把這些搜索題全A了后再回來重新寫一下，重寫效果也會不錯，并不是題目a了這題就好了。

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#include <stdio.h> #include <string.h>#define LL __int64const int mod = 100007; struct HASH{int head[mod+10], E, next[222222];int v1[222222], v2[222222], c[222222];void init() {memset(head, -1, sizeof(head));E = 0;}int gethash(int x,int y) {return (((LL)x + y)%mod+mod)%mod;}int isin(int x, int y) {int u = gethash(x, y);for(int i = head[u];i != -1;i = next[i]) {if(v1[i] == x && v2[i] == y) return c[i];}return -1;}void push(int x, int y, int step) {int u = gethash(x, y);v1[E] = x;v2[E] = y;c[E] = step;next[E] = head[u];head[u] = E++;} }ha1, ha2;struct PP{int st1, st2;int step;PP(){}PP(int st1, int st2, int step) : st1(st1), st2(st2), step(step) {} }cur, to, q[1000002];int dx[] = {1, -1, 0, 0}; int dy[] = {0, 0, 1, -1}; int a[22], b[22], d[2222];void swap(int &a, int &b) {int t = a; a = b; b = t; }void get_zt(int st1, int st2) {for(int i = 9;i >= 1;i --) a[i] = st1%10, st1 /= 10;for(int i = 9;i >= 1;i --) b[i] = st2%10, st2 /= 10; }void get_xy(int &st1, int &st2) {st1 = st2 = 0;for(int i = 1;i <= 9; i++) st1 = st1*10+a[i];for(int i = 1;i <= 9; i++) st2 = st2*10+b[i]; }bool judge(int go) {int ret = 0;for(int i = 1;i <= 9; i++) ret = ret*10+a[i];return go == ret; }int getnew(int a, int b) {int vis[5], i;for(i = 1;i <= 3; i++) vis[i] = 0;vis[a]++;vis[b]++;for(i = 1;i <= 3; i++) if(!vis[i]) return i; }int min(int a, int b) {return a>b?b:a; }int bfs(int st1,int st2, int go) {ha1.init();ha1.push(st1, st2, 0);int head = 0, tail = 0;q[tail++] = PP(st1, st2, 0);int i, j, k;while(head < tail) {cur = q[head++];st1 = cur.st1, st2 = cur.st2;get_zt(st1, st2);int now ;for(i = 1;i <= 9; i++) if(a[i] == 4) {now = i; break;}int x = now/3+1, y = now%3;if(y == 0) y = 3, x--;for(i = 0;i < 4; i++) {int xx = x+dx[i];int yy = y+dy[i];if(xx < 1 || xx > 3 || yy < 1 || yy > 3) continue;int then = (xx-1)*3+yy;swap(a[now], a[then]);swap(b[now], b[then]);if(i < 2) {swap(a[now], b[now]);if(judge(go)) {return cur.step+1;}get_xy(to.st1, to.st2);to.step = cur.step+1;if(ha1.isin(to.st1, to.st2) == -1 && to.step <= 20) {ha1.push(to.st1, to.st2, to.step);if(to.step > 20) continue;q[tail++] = to;}swap(a[now], b[now]);}else {a[now] = getnew(a[now], b[now]);if(judge(go)) {return cur.step+1;}get_xy(to.st1, to.st2);to.step = cur.step+1;if(ha1.isin(to.st1, to.st2) == -1 && to.step <= 20) {ha1.push(to.st1, to.st2, to.step);if(to.step > 20) continue;q[tail++] = to;}a[now] = getnew(a[now], b[now]);}swap(a[now], a[then]);swap(b[now], b[then]);}}int temp , tot = 0;int N = 1<<8;for(i = 0;i < N; i++) {temp = go;for(j = 9;j >= 1; j--)a[j] = temp%10 , temp /= 10;int id1 = 1, id2 = 0, sum = 0;while(id1 <= 9) {if(a[id1] == 4) {sum = sum*10+4;id1++;continue;}if((1<<id2) & i) {for(k = 3;k >= 1; k--) if(k != a[id1]) break;sum = sum*10+k;}else {for(k = 1;k <= 3; k++) if(k != a[id1]) break;sum = sum*10+k;}id1++, id2++;}d[tot++] = sum;}for(i = 0;i < tot; i++){int now = ha1.isin(go, d[i]);if(now >= 0) return now;}int ans = 1111111;for(int ii = 0;ii < tot;ii++) {ha2.init();cur = PP(go, d[ii], 0);int head = 0, tail = 0;q[tail++] = cur;int ok = 0;while(head < tail) {cur = q[head++];st1 = cur.st1, st2 = cur.st2;get_zt(st1, st2);int now ;for(i = 1;i <= 9; i++) if(a[i] == 4) {now = i; break;}int x = now/3+1, y = now%3;if(y == 0) y = 3, x--;for(i = 0;i < 4; i++) {int xx = x+dx[i];int yy = y+dy[i];if(xx < 1 || xx > 3 || yy < 1 || yy > 3) continue;int then = (xx-1)*3+yy;swap(a[now], a[then]);swap(b[now], b[then]);if(i < 2) {swap(a[now], b[now]);get_xy(to.st1, to.st2);to.step = cur.step+1;int temp = ha1.isin(to.st1, to.st2);if(temp >= 0) {ok = 1;ans = min(ans, temp+to.step);continue;}if(ha2.isin(to.st1, to.st2) == -1 && to.step <= 10) {ha2.push(to.st1, to.st2, to.step);q[tail++] = to;}swap(a[now], b[now]);}else {a[now] = getnew(a[now], b[now]);get_xy(to.st1, to.st2);to.step = cur.step+1;int temp = ha1.isin(to.st1, to.st2);if(temp >= 0) {ok = 1;ans = min(ans, temp+to.step);continue;}if(ha2.isin(to.st1, to.st2) == -1 && to.step <= 10) {ha2.push(to.st1, to.st2, to.step);q[tail++] = to;}a[now] = getnew(a[now], b[now]);}swap(a[now], a[then]);swap(b[now], b[then]);if(ok) break;}if(ok) break;}}if(ans > 30) return -1;return ans; }int main() {int i, x, n, m;while(scanf("%d%d", &m, &n) != -1 && n) {int go = 0, st1 = 0, st2 = 0;char ch[2];for(i = 1;i <= 9; i++) {scanf("%s", ch);if(ch[0] == 'W') x = 1;else if(ch[0] == 'R') x = 2;else if(ch[0] == 'B') x = 3;else x = 4;go = go*10 + x;if(i == (n-1)*3 + m) {st1 = st1*10+4;st2 = st2*10+4;}else {st1 = st1*10+1;st2 = st2*10+2;}}if(st1 == go) {puts("0"); continue;}printf("%d\n", bfs(st1, st2, go));}return 0; }

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轉載于:https://www.cnblogs.com/pangblog/p/3239133.html

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