解題區/解題區 - UVa

UVa 1558 - Number Game

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

給予 2 到 20 的內數字，兩名玩家輪流挑一個數字 x，並且將 x 的倍數遮蔽、x 倍數和已經被遮蔽的數字和也應該被遮蔽。

被遮蔽的數字將無法被使用。無法選擇任何數字的人輸。給定盤面上剩餘可選的數字，請問先手在第一步可以選擇哪幾個數字獲得勝利。

Sample Input

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5
2
1
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2
2 3

Sample Output

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5
Scenario #1:
The winning moves are: 2.
Scenario #2:
There is no winning move.

Solution

由於數字量很小，可以進行 bitmask 壓縮，來得知數字的使用情況。接著套用博弈 dp 的思路，來得知必勝狀態！

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#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[1<<20];
int pick(int state, int x) {
    for (int i = x; i <= 20; i += x)
        if ((state>>(i-2))&1)
            state ^= 1<<(i-2);
    
    for (int i = x; i <= 20; i++) {
        if ((state>>(i-2))&1) {
            if (i - x >= 2) {
                if (((state>>(i - x -2))&1) == 0)
                    state ^= 1<<(i-2);
            }
        }
    }
    return state;
}
int dfs(int state) {
    if (state == 0)	
        return 0;
    if (dp[state] != -1)	
        return dp[state];
    int &ret = dp[state];
    ret = 0;
    for (int i = 2; i <= 20; i++) {
        if ((state>>(i-2))&1) {
            int v = dfs(pick(state, i));
            ret |= !v;
            if (ret) break;
        }
    }
    return ret;
}
int main() {
    memset(dp, -1, sizeof(dp));
    int testcase, cases = 0;
    int n;
    scanf("%d", &testcase);
    while (testcase--) {
        scanf("%d", &n);
        int mask = 0, x;
        for (int i = 0; i < n; i++) {
            scanf("%d", &x);
            mask |= 1<<(x-2);
        }
        
        printf("Scenario #%d:\n", ++cases);
        vector<int> ret;
        for (int i = 2; i <= 20; i++) {
            if ((mask>>(i-2))&1) {
                int v = dfs(pick(mask, i));
                if (v == 0)
                    ret.push_back(i);
            }
        }
        if (ret.size() == 0)
            puts("There is no winning move.");
        else {
            printf("The winning moves are:");
            for (int i = 0; i < ret.size(); i++)
                printf(" %d", ret[i]);
            puts(".");
        }
        puts("");
    }
    return 0;
}