UVALive 6322 最大匹配

思路：枚举每个位置的最小字符，用最大匹配判断是否可行

#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

#define Maxn 1010

using namespace std;

int match[Maxn],vi[Maxn],num[Maxn],g[Maxn][Maxn],f[Maxn],ch[Maxn],n,le;

void init()

{

    memset(match,-1,sizeof(match));

    memset(vi,0,sizeof(vi));

    memset(num,0,sizeof(num));

    memset(g,0,sizeof(g));

    memset(f,0,sizeof(f));

    memset(ch,0,sizeof(ch));

}

int dfs(int u)

{

    int i;

    for(i=1;i<=n;i++) if(!f[i]&&!vi[i]&&g[u][i]){

        vi[i]=1;

        if(match[i]==-1||dfs(match[i])){

            match[i]=u;

            return 1;

        }

    }

    return 0;

}

int OK()

{

    int i,ans=0;

    memset(match,-1,sizeof(match));

    for(i=1;i<=n;i++){

        if(ch[i]) continue;

        memset(vi,0,sizeof(vi));

        if(dfs(i))

            ans++;

    }

    return ans==le;

}

int main()

{

    int i,j,t,l,k;

    char str[Maxn],s[Maxn];

    scanf("%d",&t);

    while(t--){

        init();

        scanf("%s",str+1);

        n=strlen(str+1);

        sort(str+1,str+1+n);

        for(i=1;i<=n;i++){

            scanf("%s",s);

            l=strlen(s);

            for(j=1;j<=n;j++){

                for(k=0;k<l;k++){

                    if(str[j]==s[k]){

                        g[i][j]=1;

                        //cout<<i<<" "<<j<<endl;

                    }

                }

            }

        }

        int ans=0;

        for(i=1;i<=n;i++){

            memset(vi,0,sizeof(vi));

            if(dfs(i))

                ans++;

        }

        if(ans<n){

            printf("NO SOLUTION\n");

            continue;

        }

        memset(match,-1,sizeof(match));

        le=n;

        for(i=1;i<=n;i++){

            for(j=1;j<=n;j++)if(!f[j]&&g[i][j]){

                ch[i]=j;

                f[j]=1;

                le--;

                if(OK())

                    break;

                le++;

                f[j]=0;

            }

        }

        for(i=1;i<=n;i++)

            printf("%c",str[ch[i]]);

        puts("");

    }

    return 0;

}