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| Description For the hope of a forever love, Steven is planning to send a ring to Jane with a romantic string engraved on. The string‘s length should not exceed N. The careful Steven knows Jane so deeply that he knows her favorite words, such as "love", "forever". Also, he knows the value of each word. The higher value a word has the more joy Jane will get when see it. Input The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case starts with a line consisting of two integers: N, M, indicating the string‘s length and the number of Jane‘s favorite words. Each of the following M lines consists of a favorite word Si. The last line of each test case consists of M integers, while the i-th number indicates the value of Si. Output For each test case, output the string to engrave on a single line. Sample Input
Sample Output
Hint Sample 1: weight(love) = 5, weight(ever) = 5, so weight(lovever) = 5 + 5 = 10 Sample 2: weight(ab) = 2 * 5 = 10, so weight(abab) = 10 |
比较普通的AC自动机+DP，和之前做的类型题差不多，这道题不同之处在于需要沿途记录字符串，并且字符串要是字典序最小的。
可以用STL里的string记录，速度感觉还行，不是很慢。
题意和思路这里就不说了，比较简单，和前面的都一样，可以参考我之前的博客。
#include<cstdio> #include<cstring> #include<queue> #include<algorithm> #include<string> #include<iostream> using namespace std; int ch[102*12][26],End[102*12],cur,fail[102*12],last[102*12],h[102]; char str[55],str0[102][12]; void get_fail() { int now,tmpFail,Next; queue<int> q; for(int j=0; j<26; j++) { if(ch[0][j]) { q.push(ch[0][j]); fail[ch[0][j]] = 0; last[ch[0][j]] = 0; } } while(!q.empty()) { now = q.front(); q.pop(); for(int j=0; j<26; j++) { if(!ch[now][j]) continue; Next = ch[now][j]; q.push(Next); tmpFail = fail[now]; while(tmpFail&&!ch[tmpFail][j]) tmpFail = fail[tmpFail]; fail[Next] = ch[tmpFail][j]; last[Next] = End[fail[Next]] ? fail[Next]:last[fail[Next]]; } } } string ans_str[102*12][55]; int dp[102*12][55]; int main() { int n,m,now,T,kase=1; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); memset(ch,0,sizeof(ch)); memset(End,0,sizeof(End)); memset(last,0,sizeof(last)); cur = 1; int len; for(int i=1; i<=m; i++) { scanf("%s",str0[i]); len = strlen(str0[i]); now = 0; for(int j=0; j<len; j++) { str0[i][j]-=‘a‘; if(ch[now][str0[i][j]]==0) ch[now][str0[i][j]] = cur++; now = ch[now][str0[i][j]]; str0[i][j]+=‘a‘; } End[now] = i; } get_fail(); for(int i=1; i<=m; i++) scanf("%d",&h[i]); memset(dp,-1,sizeof(dp)); dp[0][0]=0; ans_str[0][0]=string(""); int ans=-1,ans_now,ans_len; for(int i=0; i<n; i++) { for(int tnow=0; tnow<cur; tnow++) { if(dp[tnow][i]!=-1) for(int c=0; c<26; c++) { int now = tnow,tsum=0; while(now&&!ch[now][c]) now = fail[now]; now = ch[now][c]; if(End[now]) tsum+=h[End[now]]; int tmp = now; while(last[tmp]) { tsum += h[End[last[tmp]]]; tmp = last[tmp]; } string tmp_s = string("a"); tmp_s[0]=c+‘a‘; if(dp[tnow][i]+tsum>dp[now][i+1]|| (dp[tnow][i]+tsum==dp[now][i+1] && ans_str[tnow][i]+tmp_s < ans_str[now][i+1]) ) { ans_str[now][i+1] = ans_str[tnow][i]+tmp_s; dp[now][i+1] = dp[tnow][i]+tsum; } if(dp[now][i+1]>ans || ( dp[now][i+1]==ans && ans_len==i+1 && ( ans_str[ans_now][ans_len] > ans_str[now][i+1] ) )) { ans = dp[now][i+1]; ans_now = now; ans_len = i+1; } } } } if(ans>0) cout<<ans_str[ans_now][ans_len]; puts(""); } }
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原文地址：http://www.cnblogs.com/lastone/p/5372367.html
