Codeforces #332 B. Spongebob and Joke （思路）

本文介绍了一个关于序列匹配的问题，通过记录元素位置判断序列是否有多解、无解或可能的解，并给出了具体的解决算法及实现代码。

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题意:

$给定长度为n的序列f, 长度为m的序列b, 且f_{a_i}=b[i], 求长度为m的序列$
$多解输出Ambiguity, 无解输出Impossible, 有解输出Possible和解$

分析:

$其实f中元素的位置记录一下, 如果一个元素存在于多个位置, 这元素是Ambiguous的, -1标记下$
$输出答案的时候, 如是这个数字不存在肯定是Impossible$
$但是Ambiguity要Ambiguous的元素出现了才行, 才行, 才行。。。。$ 思维不够缜密啊！！！

数据:

3 1
2 1 1
2

ans:
Possible
1

代码:

//
//  Created by TaoSama on 2015-11-21
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, f[N], a[N], b[N], wh[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &n, &m) == 2) {
        memset(wh, 0, sizeof wh);
        memset(a, 0, sizeof a);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", f + i);
            if(wh[f[i]]) wh[f[i]] = -1;
            else wh[f[i]] = i;
        }

        bool Ambiguity = false, Impossible = false;
        for(int i = 1; i <= m; ++i) {
            scanf("%d", b + i);
            if(!wh[b[i]]) Impossible = true;
            else if(wh[b[i]] == -1) Ambiguity = true;
            a[i] = wh[b[i]];
        }
        if(Impossible) puts("Impossible");
        else if(Ambiguity) puts("Ambiguity");
        else {
            puts("Possible");
            for(int i = 1; i <= m; ++i)
                printf("%d%c", a[i], " \n"[i == m]);
        }
    }
    return 0;
}