洛谷 P3041 視頻游戲的連擊Video Game Combos

難度一般，不過這個(gè)數(shù)位DP其實(shí)應(yīng)該叫做記憶化搜索

題意：玩游戲時(shí)可以通過按鍵組合打出combo技能；然后是已知N個(gè)combo的按鍵方式，然后求K次按鍵最多可以放出的combo技能（combo技能之間可以重疊）。

思路：

在AC自動(dòng)機(jī)上如果一個(gè)點(diǎn)表示一個(gè)字符串以及它的后綴，則一個(gè)點(diǎn)可以對(duì)應(yīng)多個(gè)combo技能（這是關(guān)鍵點(diǎn)）

可以先用拓?fù)渑判蛱幚沓雒總€(gè)點(diǎn)對(duì)應(yīng)了多少個(gè)combo技能，處理時(shí)要用fail的反向邊即refail計(jì)算入度(這里應(yīng)該不用這么麻煩，build的時(shí)候就可以直接加上去了)

然后就是數(shù)位DP啦，這題沒啥坑點(diǎn)，思路簡潔

題目描述

Bessie is playing a video game! In the game, the three letters ‘A’, ‘B’, and ‘C’ are the only valid buttons. Bessie may press the buttons in any order she likes; however, there are only N distinct combos possible (1 <= N <= 20). Combo i is represented as a string S_i which has a length between 1 and 15 and contains only the letters ‘A’, ‘B’, and ‘C’.

Whenever Bessie presses a combination of letters that matches with a combo, she gets one point for the combo. Combos may overlap with each other or even finish at the same time! For example if N = 3 and the three possible combos are “ABA”, “CB”, and “ABACB”, and Bessie presses “ABACB”, she will end with 3 points. Bessie may score points for a single combo more than once.

Bessie of course wants to earn points as quickly as possible. If she presses exactly K buttons (1 <= K <= 1,000), what is the maximum number of points she can earn?

貝西在玩一款游戲，該游戲只有三個(gè)技能鍵 “A”“B”“C”可用，但這些鍵可用形成N種(1 <= N<= 20)特定的組合技。第i個(gè)組合技用一個(gè)長度為1到15的字符串S_i表示。

當(dāng)貝西輸入的一個(gè)字符序列和一個(gè)組合技匹配的時(shí)候，他將獲得1分。特殊的，他輸入的一個(gè)字符序列有可能同時(shí)和若干個(gè)組合技匹配，比如N=3時(shí)，3種組合技分別為"ABA", “CB”, 和"ABACB",若貝西輸入"ABACB"，他將獲得3分。

若貝西輸入恰好K (1 <= K <= 1,000)個(gè)字符，他最多能獲得多少分？

輸入格式

• Line 1: Two space-separated integers: N and K.

• Lines 2…N+1: Line i+1 contains only the string S_i, representing combo i.

輸出格式

• Line 1: A single integer, the maximum number of points Bessie can obtain.

輸入輸出樣例

輸入
3 7
ABA
CB
ABACB
輸出
4
說明/提示

The optimal sequence of buttons in this case is ABACBCB, which gives 4 points–1 from ABA, 1 from ABACB, and 2 from CB.

//#pragma comment(linker, "/STACK:102400000,102400000") #include "bits/stdc++.h" #define pb push_back #define ls l,m,now<<1 #define rs m+1,r,now<<1|1 #define hhh printf("hhh\n") #define see(x) (cerr<<(#x)<<'='<<(x)<<endl) using namespace std; typedef long long ll; typedef pair<int,int> pr; inline int read() {int x=0;char c=getchar();while(c<'0'||c>'9')c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return x;}const int maxn = 4e2+10; const int mod = 1e4+7; const double eps = 1e-9;char s[maxn]; int N, K; int trie[maxn][26], fail[maxn], cnt; vector<int> refail[maxn]; int in[maxn], num[maxn]; int dp[1010][maxn];void insert() {int len=strlen(s), p=0;for(int i=0; i<len; ++i) {int k=s[i]-'A';if(!trie[p][k]) trie[p][k]=++cnt;p=trie[p][k];}num[p]++; }void build() {queue<int> q;for(int i=0; i<26; ++i) if(trie[0][i]) q.push(trie[0][i]);while(q.size()) {int now=q.front(); q.pop();for(int i=0; i<26; ++i) {if(trie[now][i]) {fail[trie[now][i]]=trie[fail[now]][i];refail[trie[fail[now]][i]].pb(trie[now][i]);in[trie[now][i]]++;q.push(trie[now][i]);}else trie[now][i]=trie[fail[now]][i];}} }void Toposort() {queue<int> q;for(int i=0; i<=cnt; ++i) if(!in[i]) q.push(i);while(q.size()) {int now=q.front(); q.pop();for(int i=0; i<refail[now].size(); ++i) {int v=refail[now][i];num[v]+=num[now];if(--in[v]==0) q.push(v);}} }int dfs(int pos, int now) {if(pos>K) return 0;if(~dp[pos][now]) return dp[pos][now];int ans=0;for(int i=0; i<26; ++i)ans=max(ans,num[trie[now][i]]+dfs(pos+1,trie[now][i]));return dp[pos][now]=ans; }int main() {//ios::sync_with_stdio(false);N=read(), K=read();for(int i=0; i<N; ++i) scanf("%s", s), insert();build();Toposort();memset(dp,-1,sizeof(dp));printf("%d\n", dfs(1,0)); }

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