delete[] c;53returnlen;54}55voidPrintLCS(int**b,char*str1,inti,intj)56{57if(i==0|| j==0)58return;59if(b[i][j]==0)60{61PrintLCS(b, str1, i-1, j-1);//从后面开始递归,所以要先递归到子串的前面,然后从前往后开始输出子串62printf("%c",str1[i-1]);//c[][]的第i行元素...
1. CF#632 C.Eugene and an array(2) 2. 冒泡排序的交换次数 (树状数组)(1) 3. 匹配问题总结(1) 推荐排行榜 1. CF#632 C.Eugene and an array(2) 2. CF-1209D Cow and Snacks (并查集,图)(2) 3. Codeforces Round #670 (Div. 2) 详细题解(1) 4. 计算n个数字两两相乘的和...
代码: #include<cstdio>#include<cstring>#include<iostream>#include<string>#include<algorithm>usingnamespacestd;constintMAXN=3e2+10;constintINF=0x3f3f3f3f;chars[MAXN];chart[MAXN];intdp[MAXN][MAXN];voidLCS(intlen1,intlen2){for(inti=0;i<len1;++i){for(intj=0;j<len2;++j){if(s[...
#include<iostream>//最长公共子序列 #include<string> usingnamespacestd; stringstr1,str2; intf[500][500]; intlcs(inti,intj) { if(i==-1||j==-1) return0; if(f[i][j]!=-1) returnf[i][j]; elseif(str1[i]==str2[j]) f[i][j]=lcs(i-1,j-1)+1; else f[i][j]=max(l...
最长公共子序列求解问题。 1 #include 2 #include 3 const int maxn=1000; 4 char s1[maxn],s2[maxn]; 5 int c[maxn][maxn]; 6 int main() 7 { 8 while(scanf("%s%s&quo
For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y. InputThe program input is from...
1#include"stdio.h"23//#include "stdlib.h"45#include"string.h"67#defineN 100089chara[N],b[N];1011intc[N+2][N+2];1213intdp(intlena,intlenb)1415{1617inti,j;1819for(i=0;i<=lena;i++) {c[i][0]=0;}2021for(j=1;j<=lenb;j++) {c[0][j]=0;}2223for(i=1;i<=lena;i++...
ik > of indices of X such that for all j = 1,2,...,k, xij= zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the...
题目链接经典DP问题。由于最初if(c[i][j]>=0) return c[i][j];一句中少写一个等于号造成TLE,查了好久啊……View Code 1 #include <stdio.h> 2 #include <string.h> 3 #define MAX(
创建DP数组C[][]; 图中的空白格子需要填上相应的数字(这个数字就是c[i][j]的定义,记录的LCS的长度值)。填的规则依据递归公式,简单来说:如果横竖(i,j)对应的两个元素相等,该格子的值 = c[i-1,j-1] + 1。如果不等,取c[i-1,j] 和 c[i,j-1]的最大值。首先初始化该表: ...