链接:HDU-1053:Advanced Fruits 题意:将两个字符串合成一个串,不改变原串的相对顺序,可将相同字母合成一个,求合成后最短的字符串. 题解:LCS有三种状态转移方式,将每个点的状态转移方式记录下来,再回溯. #include <bits/stdc++.h> using namespace std; ; const int INF = 0x3f3f3f3f; ; + ; char s[maxn], t[maxn]; int slen, tlen; int dp[maxn][m…

Entropy Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5972    Accepted Submission(s): 2507 Problem Description An entropy encoder is a data encoding method that achieves lossless data compress…

#include<iostream> #include<algorithm> using namespace std; bool cmp(int a,int b) { return a>b; //从大到小排序 } int a[30],m; char s[1005]; int huffman() { int n,i,t; memset(a,0,sizeof(a)); for(i=0;i<m;i++) //统计频率 if(s[i]=='_') a[0]++; else a[…