1 4

4
2
1
1
0
0
0
0
0

2
1 2
4
3 1 2 4

0

4
4 1 2 3

2

7
5 -3 6 5 -5 -1 6
-6 1 4 -2 0 -2 0

abbbb

这是一道我连题目都看不懂的题，下面解释一下例子的意思。

初始时从0号出发，选择a0，则到达0+a0=5号点，在5号选择b5，则到达5+b5=5-2=3号点，在3号点选择b3，到达3-2=1号点，在1号点选择b1，到达1+1=2号点，在2号点选择b2，到达2+b2=6，即到达n-1号点，所以，输出的最小字典序方案为abbbb。

解题思路：首先使用一个数组G将所有可以到达t（t = 0~n-1）点的点都存起来，然后使用bfs从后向前遍历G，求得一个vis数组，若vis[0]被访问过，则说明存在一条从0到n-1的路径，否则无解。从起点0开始根据vis求字典序最小的路径，先查看a方案是否是可以到达n-1点的方案，若是，再查看a方案到达的点是否被再次访问过，若是则说明路径中存在环路，则输出“Infinity!”。同理，对方案b也一样。

下面是代码：

 public static void main(String[] args) {Scanner sc = new Scanner(System.in);while (sc.hasNext()) {int n = sc.nextInt();int[] a = new int[n];int[] b = new int[n];String[] G = new String[n];String str = "";int[] vis = new int[n];int[] vis2 = new int[n];for (int i = 0; i < n; i++)G[i] = "\t";for (int i = 0; i < n; i++) {a[i] = sc.nextInt();int t = i + a[i];if (t >= 0 && t < n) {//G存储可以到达t点的所有点（t = 0~n-1）G[t] += i + "\t";}}for (int i = 0; i < n; i++) {b[i] = sc.nextInt();int t = i + b[i];if (t >= 0 && t < n) {G[t] += i + "\t";}}//System.out.println(Arrays.toString(G));//从后向前使用广度优先搜索，求搜索路径rev_bfs(n - 1, G, vis);if (vis[0] == 0) {//从后向前搜索，未能到达初始点0System.out.println("No solution!");} else {//能够到达初始点0,则求最小字典序方案int p = 0;vis2[0] = 1; //前向访问数组boolean infflag = false;for (int x = 0; x != n - 1 && !infflag; ) {int nxt = x + a[x];if (nxt >= 0 && nxt < n && vis[nxt] == 1) {if (vis2[nxt] == 0) {vis2[nxt] = 1;str += 'a';} else {//若已访问过，说明存在环，使得方案无限infflag = true;}x = nxt;} else {nxt = x + b[x];if (nxt >= 0 && nxt < n && vis[nxt] == 1) {if (vis2[nxt] == 0) {vis2[nxt] = 1;str += 'b';} else {infflag = true;}} else {System.out.println("No solution!");}x = nxt;}}System.out.println(infflag ? "Infinity!" : str);}}}private static void rev_bfs(int p, String[] G, int[] vis) {Queue<Integer> q = new ArrayDeque<>();vis[p] = 1;q.add(p);while (!q.isEmpty()) {//System.out.println(q.toString());int x = q.poll();String[] str = G[x].trim().split("\t");for (int i = 0; i < str.length; i++) {int t;try {t = Integer.valueOf(str[i]);} catch (Exception e) {continue;}if (vis[t] == 0) {vis[t] = 1;q.add(t);}}}}

E、围棋

解题思路：根据题意模拟。这里的三个错误：miss 1 是重复放子，比较好盘判断，只要棋盘上相应位置已经存在棋子，再想要将棋子放到那里，就是miss 1。miss 2 是将棋子放在禁放位上，即，将棋子放入后，周围环绕着它的都是与它不同色的棋子，除非，将棋子放入后可以将它周围一个异色棋子给拿出来（因为这个异色棋子周围环绕的都是和它异色的棋子），用DFS检查这个棋子周围是否都是异色棋子。miss 3 是判断当前棋盘局面是否和历史棋盘局面一样，可以利用hash判断当前棋盘局面是否出现过。

package CodeM;import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;class hascode {long has0;long has1;public hascode(long has0, long has1) {this.has0 = has0;this.has1 = has1;}
}public class GameGo_2017 {static int[][] dir = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};static int goNum = 19;public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();while (n-- > 0) {int Num = sc.nextInt();char[][] go = new char[goNum][goNum];for (int i = 0; i < go.length; i++) {// inital go boardfor (int j = 0; j < go.length; j++)go[i][j] = '.';}//use st to store the hashcode of go board,// to compare the next go board and the past go board.ArrayList<Long> st = new ArrayList<>();check(go, st);while (Num-- > 0) {String op = sc.next();int x = sc.nextInt();int y = sc.nextInt();x -= 1;y -= 1;//System.out.println("first : "+op);char a = op.charAt(0);char b = a == 'B' ? 'W' : 'B';if (go[x][y] != '.') {//exist one point.System.out.println("miss 1");} else {//use nextgo to get the next step//if this step can be done then cur equals to nextgo//otherwise print the mistake.char[][] nextgo = new char[goNum][goNum];copy(nextgo, go);/*System.out.println("==============================");for (int i = 0; i < goNum; i++) {System.out.println(Arrays.toString(nextgo[i]));}*/nextgo[x][y] = a;//use vis[][] to distinguish one poist is visited or not.int[][] vis = new int[goNum][goNum];//check this step is conformed to the 2nd or not//that is whether exist a or not,if not that is the 2nd mistake.boolean isOk = dfs(x, y, a, vis, nextgo);for (int i = 0; i < dir.length; i++) {int tx = x + dir[i][0];int ty = y + dir[i][1];//!dfs(tx,ty,b,vis,nextgo) means there are all a around [tx,ty]//according the go rules,must pick the go point.if (tx >= 0 && tx < goNum && ty >= 0 && ty < goNum &&vis[tx][ty] == 0 && nextgo[tx][ty] == b&& !dfs(tx, ty, b, vis, nextgo)) {pick(tx, ty, b, nextgo);isOk = true;}}if (isOk) {// if this step will not appear the first two mistakes// now check if the same as the the past one go board.// use hashcode to compare.if (check(nextgo, st)) {//if this step is pass the check,then the cur is// the same as the nextcopy(go, nextgo);} else System.out.println("miss 3");} else System.out.println("miss 2" );}}//show();for (int i = 0; i < goNum; i++) {for (int j = 0; j < goNum; j++) {System.out.print(go[i][j]);}System.out.println();}}}private static boolean dfs(int x, int y, char op, int[][] vis, char[][] next) {//check whether this point is surrounded by different pointsif (next[x][y] == '.') return true;if (next[x][y] != op) return false;vis[x][y] = 1;boolean isOk = false;for (int i = 0; i < dir.length; i++) {int tx = x + dir[i][0];int ty = y + dir[i][1];if (tx >= 0 && tx < goNum && ty >= 0 && ty < goNum && vis[tx][ty] == 0) {isOk |= dfs(tx, ty, op, vis, next);}}return isOk;}private static void pick(int x, int y, char op, char[][] next) {next[x][y] = '.';for (int i = 0; i < dir.length; i++) {int tx = x + dir[i][0];int ty = y + dir[i][1];if (tx >= 0 && tx < goNum && ty >= 0 && ty < goNum && next[tx][ty] == op)pick(tx, ty, op, next);}}private static boolean check1(char[][] next, ArrayList<hascode> st) {long has1 = 0, has2 = 0;for (int i = 0; i < goNum; i++) {for (int j = 0; j < goNum; j++) {has1 = has1 * 131 + next[i][j];has2 = has2 * 137 + next[i][j];}}for (hascode has : st) {if (has1 == has.has0 && has2 == has.has1)return false;}hascode hasCode = new hascode(has1, has2);st.add(hasCode);return true;}private static boolean check(char[][] next, ArrayList<Long> st) {long has1 = 0;for (int i = 0; i < goNum; i++) {for (int j = 0; j < goNum; j++) {has1 = has1 * 131 + next[i][j];//has2 = has2 * 137 + next[i][j];}}if (st.contains(has1)){return false;}//hascode hasCode = new hascode(has1, has2);st.add(has1);return true;}private static void copy(char[][] next, char[][] go) {for (int i = 0; i < goNum; i++)for (int j = 0; j < goNum; j++)next[i][j] = go[i][j];}}