前言

MOOK程序设计与算法（二）算法基础-郭伟-----学习笔记

数字三角形 The Triangle（poj1163）

Description

7
3 8
8 1 0
2 7 4 4
4 5 2 6 5

(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.
题目大意：求从上到下最长的路径，没次可以沿对角线向左或向右走；

递归解法

#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int Max = 105;
int N = 0;
int a[105][105];
int tem[Max][Max];
int max_d(int i,int j){
	if(tem[i][j]!=0) return tem[i][j];
	if(i==N) return a[i][j];
	return tem[i][j]=max(max_d(i+1,j),max_d(i+1,j+1))+a[i][j];
}
int main(){
	cin>>N;
	for(int i = 1;i<=N;i++){
		for(int j = 1;j<=i;j++){
			cin>>a[i][j];
		}
	}
	cout<<max_d(1,1);
	return 0;
}

递推解法

#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int Max = 105;
int N = 0;
int a[105][105];
//int max_d(int i,int j){
//	if(tem[i][j]!=0) return tem[i][j];
//	if(i==N) return a[i][j];
//	return tem[i][j]=max(max_d(i+1,j),max_d(i+1,j+1))+a[i][j];
//}
int main(){
	cin>>N;
	for(int i = 1;i<=N;i++){
		for(int j = 1;j<=i;j++){
			cin>>a[i][j];
		}
	}
	
	for(int i = N-1;i>=1;i--){
		for(int j = 1;j<=i;j++){
			a[N][j] = max(a[N][j],a[N][j+1])+a[i][j];
		}	
	}
     cout<<a[N][1];
	return 0;
}

最长上升子序列LIS（百练2757）

题目：
一个数的序列bi，当b1 < b2 < … < bS的时候，我们称这个序列是上升的。对于给定的一个序列(a1, a2, …, aN)，我们可以得到一些上升的子序列(ai1, ai2, …, aiK)，这里1 <= i1 < i2 < … < iK <= N。比如，对于序列(1, 7, 3, 5, 9, 4, 8)，有它的一些上升子序列，如(1, 7), (3, 4, 8)等等。这些子序列中最长的长度是4，比如子序列(1, 3, 5, 8).
你的任务，就是对于给定的序列，求出最长上升子序列的长度。

题解

#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int n = 10005;
int a[n];
int maxLen[n];
int main(){
	int N = 0;
	cin>>N;
	for(int i = 1;i<=N;i++){
		cin>>a[i];
		maxLen[i] = 1;
	}
	for(int i = 2;i<=N;i++ ){
		for(int j = 1;j<=i;j++){
			if(a[j]<a[i])
			maxLen[i] = max(maxLen[i],maxLen[j]+1);
		}
	}
	cout<<*max_element(maxLen+1,maxLen+N+1);
	return 0;
}

最长公共子序列LCS（poj458）

题目：
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, …, xm > another sequence Z = < z1, z2, …, zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, …, 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 maximum-length common subsequence of X and Y.

个人题解

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

typedef long long ll;

string str1,str2;  
int dp[1000][1000];

int main(){
	while(cin>>str1>>str2){
	memset(dp,0,sizeof(dp));
	for(int i = 1;i<=str1.length();i++){
		for(int j = 1;j<=str2.length();j++){
			if(str1[i-1]==str2[j-1])
			dp[i][j] = dp[i-1][j-1]+1;
			else 
			dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
		}
	}
		cout<< dp[str1.length()][str2.length()]<<endl;
	}
	return 0;
}//自己测都没问题，提交后都编译错误