hihoCoder [Offer收割]编程练习赛35 缩写命名

题目链接http://hihocoder.com/contest/offers35/problem/4

题意:略。

思路:对于原串中的每个位置的字符,如果词典中有字符串有这个位置的字符,那么就由这个位置向这个词典中的字符串连一条边,题目要求要找到一种方案使得每个位置的字符都有对应的一个词典中的字符串,那么问题就转化为了经典的二分匹配,连完边后跑一下匈牙利算法看最后匹配数即可知道结果。

#pragma comment(linker, "/STACK:102400000,102400000")
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define Key_Value ch[ch[root][1]][0]
#define DBN1(a)           cerr<<#a<<"="<<(a)<<"\n"
#define DBN2(a,b)         cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n"
#define DBN3(a,b,c)       cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define DBN4(a,b,c,d)     cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"
#define DBN5(a,b,c,d,e)   cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n"
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=500000+5;
const int INF=0x3f3f3f3f;
const int P=1000000007;
const double PI=acos(-1.0);
template<typename T>
inline T read(T&x){
    x=0;int _f=0;char ch=getchar();
    while(ch<'0'||ch>'9')_f|=(ch=='-'),ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x=_f?-x:x;
}
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
int T,n,pre[105];
bool match[105][105],used[105];
string s,t;
bool dfs(int x){
    for (int i=1;i<=n;i++){
        if (match[x][i] && !used[i]){
            used[i]=true;
            if (pre[i]==0 || dfs(pre[i])){
                pre[i]=x;
                return true;
            }
        }
    }
    return false;
}
int main(){
    for (read(T);T--;){
        read(n);
        cin>>s;
        memset(match,false,sizeof(match));
        memset(pre,0,sizeof(pre));
        for (int i=1;i<=n;i++){
            cin>>t;
            for (int j=0;j<(int)s.length();j++){
                for (int k=0;k<(int)t.length();k++){
                    if (s[j]==t[k]){
                        match[j+1][i]=true;
                        break;
                    }
                }
            }
        }
        int ans=0;
        for (int i=1;i<=s.length();i++){
            memset(used,false,sizeof(used));
            if (dfs(i)) ans++;
        }
        puts(ans==s.length()?"Yes":"No");
    }
    return 0;
}

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