UVa #10003 Cutting Sticks (例题9-9)

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9-8还没做完，感觉有点困难。回头再做

这道题很好的训练了最优矩阵连乘。记忆化搜索TLE了，递推则很顺畅。

递推的顺序：不能按照以前的方法递推，因为求某一个区间的解的时候要用到它子区间的解，如果按照以前的顺序递推，则子区间的解此时还没有求出来。所以先把长度为1的子区间的解求出来，这样就能求长度为2的子区间的解，然后就能求3、4...直到n+1

Run Time: 0.135s

#define UVa  "LT9-9.10003.cpp"//Cutting Stickschar fileIn[30] = UVa, fileOut[30] = UVa;#include<cstring>#include<cstdio>#include<algorithm>using namespace std;//Global Variables. Reset upon Each Case!const int maxn = 50 + 10, maxl = 1000 + 10;int l, n;int c[maxn];int d[maxl][maxl];/////int main() {    while(scanf("%d", &l) && l) {        scanf("%d", &n);        for(int i = 1; i <= n; i ++) scanf("%d", &c[i]);        c[0] = 0;        c[n+1] = l;        for(int len = 1; len <= n+1; len ++) {            for(int i = 0; i+len <= n+1; i ++) {                int j = i + len;                if(len == 1)                    d[i][j] = 0;                else {                    d[i][j] = d[i][i+1] + d[i+1][j];                    for(int k = i + 2; k < j; k ++)                        d[i][j] = min(d[i][j], d[i][k]+d[k][j]);                    d[i][j] += c[j] - c[i];                }            }        }        printf("The minimum cutting is %d.\n", d[0][n+1]);    }    return 0;}//Memoization, TLE://int dp(int i, int j) {//    if(d[i][j] != -1) return d[i][j];//    if(i >= j-1) d[i][j] = 0;//    else {//        int w = c[j] - c[i];//        d[i][j] = dp(i,i+1) + dp(i+1,j) + w;//        for(int k = i+2; k < j; k ++)//            d[i][j] = min(d[i][j], dp(i,k)+dp(k,j)+w);//    }//    return d[i][j];//}

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