pku 3356 AGTC
#include
#include
Let x and y be two strings over some finite alphabet A. We would like to transform x int
\include <iostream> \include <cstdio> \include <cstring> using namespace std; int map\[3
\include <iostream> \include <string> using namespace std; int v\[1001\]\[1001\]; int ma
\include <iostream> using namespace std; int map\[100\]\[100\]; int v\[100\]\[100\]; int
/\ //很很很经典的bfs, 这也是我第一次正式用bfs来刷题, 感受颇深: //one.cpp是用bfs来解决从一个点到另一个点的最短路程长度的方法(写的有点冗杂) 注意的
\include <iostream> using namespace std; \define MAX 52 int map\[MAX\]\[MAX\]; bool visi
题意: 给定原字符串和目标字符串,有三种操作方法,转变、添加、删除字符;求得到目标字符串的最小操作数。 如题目中的示例最小操作数为4,不过这一题在poj上的测试数据比较水,
AGTC <table> <tbody> <tr> <td><strong>Time Limit:</strong> 1000MS</td>
\include<stdio.h> int str\[351\]\[351\]; int main() \{ int n,i,j; // freopen("
Let x and y be two strings over some finite alphabet A. We would like to transform xinto
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