UVA 10003 Cutting Sticks search memory segment DP +

UVA 10003 Cutting Sticks + interval DP

Even if they have played blast

Subject to the effect

Has a length L sticks, sticks with n intermediate point of tangency. The cost of each cut length of the current wooden sticks. Seeking the minimum cost of cutting sticks

input Output

The first line is a stick length L, the second row is the number of cutting points n, n rows are the coordinates of the next point on the cutting stick.

The minimum cost of a wooden cutting output

Before then - simple entry interval dp

Society interval dp following blog entry written by very good, I just look at their blog, simple entry, after application to rely on himself.

https://blog.csdn.net/qq_41661809/article/details/81487613 read, Basics

https://blog.csdn.net/qq_40772692/article/details/80183248 slightly more advanced

https://www.cnblogs.com/HDUjackyan/p/9123199.html Featured Training + code implementation

Problem-solving ideas

1. The use of dynamic programming, the definition of \ (dp [\ i \] [\ j \] \) is cut sticks \ (ij \) the optimal cost, and then use the interval to be processed dp
2. Or use the search algorithm, but in order to reduce complexity, typically using memory search.

Code

// 动态规划解法
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

const int inf=0x3f3f3f3f;
int cut[100];
int dp[100][100];

int main()
{
    int L, n;
    while(scanf("%d",&L) && L)
    {
        memset(dp, inf, sizeof(dp));
        cin>>n;
        for(int i=1; i<=n; i++){
            cin>>cut[i];
        }
        cut[0]=0;
        cut[n+1]=L;
        for(int i=0; i<=n; i++)
            dp[i][i+1]=0; //相邻两个切割点之间是一段木棍，根据dp的含义，它们的值为零
        for(int len=2; len<=n+1; len++) //枚举区间长度，从长度为2开始
        {
            for(int i=0; i+len<=n+1; i++) //起点要从0开始
            {
                int j=i+len; //终点
                for(int k=i+1; k<j; k++) //枚举断点
                {
                    dp[i][j]=min(dp[i][k]+dp[k][j]+cut[j]-cut[i], dp[i][j]);
                }
            }
        }
        cout<<"The minimum cutting is "<<dp[0][n+1]<<"."<<endl;
    }
    return 0;
}

//记忆化搜索  搜索真万能！！！
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1010;
const int INF = 0x3f3f3f3f;
int c[N] ,dp[N][N];
int len , n;
int d(int l, int r) {
    if(r - l == 1) {
        return 0;
    }
    if(dp[l][r] > 0) {
        return dp[l][r]; //算过>0，就不要重新再算了
    }
    dp[l][r] = INF;
    for(int i = l; i <= r; i++) {
        dp[l][r] = min( dp[l][r], d(l,i)+d(i,r) + c[r] - c[l]);
    }
    return dp[l][r];
}
int main() {
    while(scanf("%d",&len) != EOF && len) {
        scanf("%d",&n);
        for(int i = 1; i <= n; i++) {
            scanf("%d",&c[i]);
        }
        c[0] = 0;
        c[n+1] = len;
        memset(dp,0,sizeof(dp));
        int ans = d(0, n+1);
        printf("The minimum cutting is %d.\n",ans);
    }
    return 0;
}
//--------------------- 
//作者：HelloWorld10086 
//来源：CSDN 
//原文：https://blog.csdn.net/HelloWorld10086/article/details/40626461 
//版权声明：本文为博主原创文章，转载请附上博文链接！