UVA - 10003 Cutting Sticks(切木棍)（dp）

题意：有一根长度为L(L<1000)的棍子，还有n(n < 50)个切割点的位置(按照从小到大排列)。你的任务是在这些切割点的位置处把棍子切成n+1部分，使得总切割费用最小。每次切割的费用等于被切割的木棍长度。

分析：

1、solve(i, j)为切割小木棍i~j的最优费用。

2、设k(i<k<j)，solve(i, j) = min{solve(i, k) + solve(k, j)} + a[j] - a[i]。

k是切割小木棍i~j费用最优的切割点，a[j] - a[i] 为小木棍i ~ j 的长度，即切割小木棍i~j的费用。

3、将小木棍的n个切割点标记为1~n，左端点为0，右端点为n + 1，则答案为solve(0, n + 1)。

4、注意L和n个切割点的位置都是positive number，所以可能是浮点数。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 50 + 10;
const int MAXT = 10000 + 10;
using namespace std;
double a[MAXN], dp[MAXN][MAXN];
double solve(int l, int r){
    if(dcmp(dp[l][r], -1)) return dp[l][r];
    if(r - l == 1) return dp[l][r] = 0.0;
    double mi = 1e15;
    for(int k = l + 1; k < r; ++k){
        double tmp = solve(l, k) + solve(k, r);
        if(dcmp(tmp, mi) == -1) mi = tmp;
    }
    return dp[l][r] = mi + a[r] - a[l];
}
int main(){
    double L;
    while(scanf("%lf", &L) == 1){
        if(!dcmp(L, 0)) return 0;
        int n;
        scanf("%d", &n);
        a[0] = 0.0;
        for(int i = 1; i <= n; ++i){
            scanf("%lf", &a[i]);
        }
        a[n + 1] = L;
        for(int i = 0; i <= n + 1; ++i)
            for(int j = 0; j <= n + 1; ++j)
                dp[i][j] = -1;
        printf("The minimum cutting is %g.\n", solve(0, n + 1));
    }
    return 0;
}