PAT T1004 To Buy or Not to Buy - Hard Version

暴力搜索加剪枝~

#include<bits/stdc++.h>
using namespace std;
const int maxn=1014;
string t;
string s[maxn];
int pos[maxn],pos1[maxn];
int N;
int nowLength;
int minLength=1e9;
int tle;
void dfs (int v,int cnt) {
    if (tle==101) return;
    tle++;
    if (nowLength>minLength) return;
    if (cnt>=t.length()) {
        minLength=min(minLength,nowLength);
        return;
    }
    v++;
    while (v<=N) {
        int a=cnt;
        int pos2[maxn];
        for (int i=0;i<s[v].length();i++) pos2[s[v][i]]=pos1[s[v][i]];
        for (int i=0;i<s[v].length();i++) 
        if (pos[s[v][i]]>pos1[s[v][i]]) cnt++,pos1[s[v][i]]++;
        nowLength+=s[v].length();
        dfs (v,cnt);
        cnt=a;
        for (int i=0;i<s[v].length();i++) pos1[s[v][i]]=pos2[s[v][i]];
        nowLength-=s[v].length();
        v++;
    }
}
int main () {
    cin>>t>>N;
    for (int i=1;i<=N;i++) cin>>s[i];
    for (int i=0;i<t.length();i++) pos[t[i]]++;
    dfs (0,0);
    int ans=0;
    for (int i=1;i<=N;i++)
    for (int j=0;j<s[i].length();j++)
    if (pos[s[i][j]]) pos[s[i][j]]--;
    for (int i=0;i<t.length();i++) ans+=pos[t[i]],pos[t[i]]=0;
    if (minLength<1e9) printf ("Yes %d",minLength-t.length());
    else printf ("No %d",ans);
    return 0;
}