Codeforces 626B Cards 【BFS or 讨论】-优快云博客

本博客探讨了一个涉及B、G、R三种颜色卡片的博弈问题，参与者通过特定规则交换卡片直到只剩一张，分析并预测最终卡片的颜色。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Catherine has a deck of n cards, each of which is either red, green, or blue. As long as there are at least two cards left, she can do one of two actions:

take any two (not necessarily adjacent) cards with different colors and exchange them for a new card of the third color;
take any two (not necessarily adjacent) cards with the same color and exchange them for a new card with that color.

She repeats this process until there is only one card left. What are the possible colors for the final card?

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200) — the total number of cards.

The next line contains a string s of length n — the colors of the cards. s contains only the characters 'B', 'G', and 'R', representing blue, green, and red, respectively.

Output

Print a single string of up to three characters — the possible colors of the final card (using the same symbols as the input) in alphabetical order.

Examples

input
2
RB

output
G

input
3
GRG

output
BR

input
5
BBBBB

output
B

Note

In the first sample, Catherine has one red card and one blue card, which she must exchange for a green card.

In the second sample, Catherine has two green cards and one red card. She has two options: she can exchange the two green cards for a green card, then exchange the new green card and the red card for a blue card. Alternatively, she can exchange a green and a red card for a blue card, then exchange the blue card and remaining green card for a red card.

In the third sample, Catherine only has blue cards, so she can only exchange them for more blue cards.

题意：有若干个R、B、G颜色的球，你可以选择两个颜色不一样的球换成一个其他颜色的球，或者使用两个颜色相同的球，换成一个该颜色的球。问你最后剩余的球可能是哪种颜色。

思路：BFS 或者讨论。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF (2000000000+10)
#define eps 1e-8
#define MAXN (60000+10)
#define MAXM (600000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
char str[210];
struct Node{
    int b, g, r;
};
bool vis[210][210][210]; bool B, G, R;
bool judge(Node a)
{
    if(vis[a.b][a.g][a.r]) return false;
    vis[a.b][a.g][a.r] = true;
    return true;
}
void BFS(int sa, int sb, int sc)
{
    Node now, next; queue<Node> Q;
    now.b = sa, now.g = sb, now.r = sc; Q.push(now);
    while(!Q.empty())
    {
        now = Q.front(); Q.pop();
        if((now.b == 0) + (now.g == 0) + (now.r == 0) > 1)
        {
            if(now.b) B = true;
            if(now.g) G = true;
            if(now.r) R = true;
            continue;
        }
        if(now.b == 0){next = now; next.b++; next.g--; next.r--; if(judge(next)) Q.push(next);}
        else if(now.g == 0){next = now; next.g++; next.b--; next.r--; if(judge(next)) Q.push(next);}
        else if(now.r == 0){next = now; next.r++; next.b--; next.g--; if(judge(next)) Q.push(next);}
        else
        {
            next = now; next.b++; next.g--; next.r--; if(judge(next)) Q.push(next);
            next = now; next.g++; next.b--; next.r--; if(judge(next)) Q.push(next);
            next = now; next.r++; next.b--; next.g--; if(judge(next)) Q.push(next);
        }
        if(now.b == 0)
        {
            next = now; if(next.g > 1) {next.g--; if(judge(next)) Q.push(next);}
            next = now; if(next.r > 1) {next.r--; if(judge(next)) Q.push(next);}
        }
        else if(now.g == 0)
        {
            next = now; if(next.b > 1) {next.b--; if(judge(next)) Q.push(next);}
            next = now; if(next.r > 1) {next.r--; if(judge(next)) Q.push(next);}
        }
        else if(now.r == 0)
        {
            next = now; if(next.g > 1) {next.g--; if(judge(next)) Q.push(next);}
            next = now; if(next.b > 1) {next.b--; if(judge(next)) Q.push(next);}
        }
        else
        {
            next = now; if(next.b > 1) {next.b--; if(judge(next)) Q.push(next);}
            next = now; if(next.g > 1) {next.g--; if(judge(next)) Q.push(next);}
            next = now; if(next.r > 1) {next.r--; if(judge(next)) Q.push(next);}
        }
    }
}
int main()
{
    int n; Ri(n);
    Rs(str); B = G = R = false;
    int numb = 0, numg = 0, numr = 0;
    for(int i = 0; i < n; i++)
    {
        numb += (str[i] == 'B');
        numg += (str[i] == 'G');
        numr += (str[i] == 'R');
    }
    CLR(vis, false);
    BFS(numb, numg, numr);
    if(B) printf("B");
    if(G) printf("G");
    if(R) printf("R");
    printf("\n");
    return 0;
}

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF (2000000000+10)
#define eps 1e-8
#define MAXN (60000+10)
#define MAXM (600000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
char str[210];
int main()
{
    int n; Ri(n); bool B, G, R;
    Rs(str); B = G = R = false;
    int numb = 0, numg = 0, numr = 0;
    for(int i = 0; i < n; i++)
    {
        numb += (str[i] == 'B');
        numg += (str[i] == 'G');
        numr += (str[i] == 'R');
    }
    int v = (numb == 0) + (numg == 0) + (numr == 0);
    if(v == 1)
    {
        if(numb == 0)
        {
            B = true;
            if(numg == numr && numg == 1) {}
            else if(numg == 1 || numr == 1)
            {
                G = numg == 1;
                R = numr == 1;
            }
            else
            {
                G = true;
                R = true;
            }
        }
        else if(numg == 0)
        {
            G = true;
            if(numb == numr && numb == 1) {}
            else if(numb == 1 || numr == 1)
            {
                B = numb == 1;
                R = numr == 1;
            }
            else
            {
                B = true;
                R = true;
            }
        }
        else if(numr == 0)
        {
            R = true;
            if(numb == numg && numg == 1) {}
            else if(numb == 1 || numg == 1)
            {
                G = numg == 1;
                B = numb == 1;
            }
            else
            {
                G = true;
                B = true;
            }
        }
    }
    else if(v == 2)
    {
        B = numb > 0;
        G = numg > 0;
        R = numr > 0;
    }
    else{B = G = R = true;}
    if(B) printf("B");
    if(G) printf("G");
    if(R) printf("R");
    printf("\n");
    return 0;
}