动态规划（dp）入门解题报告

基本思想与策略:

    基本思想与分治法类似，也是将待求解的问题分解为若干个子问题（阶段），按顺序求解子阶段，前一子问题的解，为后一子问题的求解提供了有用的信息。在求解任一子问题时，列出各种可能的局部解，通过决策保留那些有可能达到最优的局部解，丢弃其他局部解。依次解决各子问题，最后一个子问题就是初始问题的解。

    由于动态规划解决的问题多数有重叠子问题这个特点，为减少重复计算，对每一个子问题只解一次，将其不同阶段的不同状态保存在一个二维数组中。

状态抓换方程：

         dp[i][j] = a[i][j] + max(dp[i+1][j], dp[i+1][j+1]);  //dp[i][j]表示从(i,j)位置开始走的最大和

代码：

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff											//INT_MAX
#define inf 0x3f3f3f3f    										//int范围内可靠的无穷大
#define FOR(i,a) for((i)=0;i<(a);(i)++)							                //[i,a);
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
#define LL __int64
const double PI = acos(-1.0);
template<class T> T gcd(T a, T b) {return b ? gcd(b, a % b) : a;}
template<class T> T lcm(T a, T b) {return a / gcd(a, b) * b;}
template<class T> inline T Min(T a, T b) {return a < b ? a : b;}
template<class T> inline T Max(T a, T b) {return a > b ? a : b;}
using namespace std;
int n, m;
#define N 100
int a[N][N];
int dp[N][N];
void Into(){
	sf(n);
	memset(dp,-1,sizeof(dp));
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= i; j++)
			sf(a[i][j]);
}
int slove(int i,int j){
	if(dp[i][j] >= 0)
		return dp[i][j];                             //记忆化搜索
	return dp[i][j] = a[i][j] + (i == n ? 0 : Max(slove(i + 1 , j) , slove(i + 1, j + 1)));
}
int main(){
#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
#endif
	int t;
	sf(t);
	while(t--){
		Into();
		pf(slove(1,1));
	}
	return 0;
}

第二题：HDU  2602（01背包问题）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 38897    Accepted Submission(s): 16122

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

 

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output