Cutting Sticks

题意:

l长的木棒,给出n个切割点,每切一次的费用为切得木棒的长度,完成切割的最小费用。

分析:

区间dp入门,区间dp的特点,一个大区间的解可以转换成小区间的解组合起来,每个切割点的标号代表边界。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  1000000007;
int dp[100][100],c[100],n,l;
void solve(){
    c[0]=0;
    c[n+1]=l;
    memset(dp,0,sizeof(dp));
    for(int p=2;p<=n+1;++p)
        for(int i=0;i<=n-p+1;++i)
    {
        int j=i+p;
        dp[i][j]=INF;
        for(int k=i+1;k<j;++k)
            dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+c[j]-c[i]);
    }
    printf("The minimum cutting is %d.\n",dp[0][n+1]);
}
int main()
{
    while(~scanf("%d",&l)){
        if(l==0)break;
        scanf("%d",&n);
        for(int i=1;i<=n;++i)
            scanf("%d",&c[i]);
        solve();
    }
return 0;
}

 

posted on 2015-10-11 21:49  积跬步、至千里  阅读(155)  评论(0编辑  收藏  举报

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