3
UPDATE：我发现问题是我的DP解决方案没有正确处理奖金。我向状态数组添加了一个维度来表示前6个类别的总和。但是，解决方案已超时。由于每台测试用例都可以在我的机器上少于1秒的时间解决，所以不会超时。Uva Judge 10149，Yahtzee
问题的描述是在这里：http://uva.onlinejudge.org/external/101/10149.html
我在网上搜索，发现它应该由DP和掩码来解决。我实现了代码并通过了我测试过的所有测试用例，但Uva Judge返回了错误的答案。
我的想法是让状态[i] [j]匹配第i轮到由j掩蔽的类别。请指出我的错误或链接一些可以正确解决此问题的代码。这里是我的代码：
public class P10149 {
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(new FileInputStream("input.txt"));
// Scanner in = new Scanner(System.in);
while (in.hasNextLine()) {
int[][] round = new int[13][5];
for (int i = 0; i < 13; i++) {
for (int j = 0; j < 5; j++) {
round[i][j] = in.nextInt();
}
}
in.nextLine();
int[][] point = new int[13][13];
for (int i = 0; i < 13; i++) {
for (int j = 0; j < 13; j++) {
point[i][j] = getPoint(round[i], j);
}
}
int[][] state = new int[14][1 << 13];
for (int i = 1; i <= 13; i++) {
Arrays.fill(state[i], -1);
}
int[][] bonusSum = new int[14][1 << 13];
int[][] choice = new int[14][1 << 13];
for (int i = 1; i <= 13; i++) {
for (int j = 0; j < (1 << 13); j++) {
int usedSlot = 0;
for (int b = 0; b < 13; b++) {
if (((1 << b) & j) != 0) {
usedSlot++;
}
}
if (usedSlot != i) {
continue;
}
for (int b = 0; b < 13; b++) {
if (((1 << b) & j) != 0) {
int j2 = (~(1 << b) & j);
int bonus;
if (b < 6) {
bonus = bonusSum[i - 1][j2] + point[i - 1][b];
} else {
bonus = bonusSum[i - 1][j2];
}
int newPoint;
if (bonus >= 63 && bonusSum[i - 1][j2] < 63) {
newPoint = 35 + state[i - 1][j2] + point[i - 1][b];
} else {
newPoint = state[i - 1][j2] + point[i - 1][b];
}
if (newPoint > state[i][j]) {
choice[i][j] = b;
state[i][j] = newPoint;
bonusSum[i][j] = bonus;
}
}
}
}
}
int index = (1 << 13) - 1;
int maxPoint = state[13][index];
boolean bonus = (bonusSum[13][index] >= 63);
int[] mapping = new int[13];
for (int i = 13; i >= 1; i--) {
mapping[choice[i][index]] = i;
index = (~(1 << choice[i][index]) & index);
}
for (int i = 0; i < 13; i++) {
System.out.print(point[mapping[i] - 1][i] + " ");
}
if (bonus) {
System.out.print("35 ");
} else {
System.out.print("0 ");
}
System.out.println(maxPoint);
}
}
static int getPoint(int[] round, int category) {
if (category < 6) {
int sum = 0;
for (int i = 0; i < round.length; i++) {
if (round[i] == category + 1) {
sum += category + 1;
}
}
return sum;
}
int sum = 0;
int[] count = new int[7];
for (int i = 0; i < round.length; i++) {
sum += round[i];
count[round[i]]++;
}
if (category == 6) {
return sum;
} else if (category == 7) {
for (int i = 1; i <= 6; i++) {
if (count[i] >= 3) {
return sum;
}
}
} else if (category == 8) {
for (int i = 1; i <= 6; i++) {
if (count[i] >= 4) {
return sum;
}
}
} else if (category == 9) {
for (int i = 1; i <= 6; i++) {
if (count[i] >= 5) {
return 50;
}
}
} else if (category == 10) {
for (int i = 1; i <= 3; i++) {
if (isStraight(count, i, 4)) {
return 25;
}
}
} else if (category == 11) {
for (int i = 1; i <= 2; i++) {
if (isStraight(count, i, 5)) {
return 35;
}
}
} else if (category == 12) {
for (int i = 1; i <= 6; i++) {
for (int j = 1; j <= 6; j++) {
if (i != j && count[i] == 3 && count[j] == 2) {
return 40;
}
}
}
}
return 0;
}
static boolean isStraight(int[] count, int start, int num) {
for (int i = start; i < start + num; i++) {
if (count[i] == 0) {
return false;
}
}
return true;
}
}
多少测试用例你测试？您应该始终记住测试边缘情况，如空输入，最大尺寸输入等。还要确保您的输出完全符合机器人的要求 - 一个单独的空格或小数点可以为您提供WA。 – hugomg 2011-06-01 16:23:19
还要确保您的测试输入完全采用机器人给出的格式。你的程序能否很好地处理以换行符结尾的输入？ – xofon 2011-06-02 08:35:55
输入应该没问题，否则我应该得到运行时错误而不是WA。上面的代码中有一个问题，即五种也应该被视为满屋。但即使修正后，我仍然得到一个西澳。 – 2011-06-03 11:21:19