*UVa 10003 Cutting Sticks(区间dp 切木棍)

UVA - 10003

一道区间dp变形,根据紫书的思路 : 设d(i, j)代表切割i ~ j 的最优费用,则d(i, j) = min{d(i,k) + d(k,j) | i < k < j} + a[j] - a[i],其中a[j] - a[i]代表第一刀的费用

相当于把每一刀的位置离散化处理,将木棒视为1-n每一个点都可以切割,且最终答案为dp[0][n+1],巧妙地转化成了区间dp,并且是“倒着解决问题”,将第一刀(cost = len)变成了最后一刀,或者说随着状态变化而变化,反正这一刀迟早都要切

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
int dp[55][55];
int main()
{
    int len, n;
    int a[55];
    while(scanf("%d",&len) != EOF && len)
    {
        scanf("%d",&n);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i ++)
            scanf("%d",&a[i]);
        a[0] = 0;
        a[n + 1] = len;
        for(int l = 1; l <= n + 1; l ++)//区间长度
            for(int i = 0; i + l <= n + 1; i ++)
            {
                int j = i + l;

                if(j - i >= 2){
                    dp[i][j] = inf;
                    for(int k = i + 1; k < j; k ++)
                    {
                        dp[i][j] = min(dp[i][k] + dp[k][j] + a[j] - a[i], dp[i][j]);
                    }
            }
                //if(flag) dp[i][j] = j - i;
                //flag = 1;
            }
        printf("The minimum cutting is %d.\n",dp[0][n+1]);
    }
    return 0;
}

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转载自blog.csdn.net/qq_41444888/article/details/81366031