// uva348 最优矩阵链乘// 典型的区间dp// dp[i][j] 表示矩阵i到j链乘所得到的最小花费// dp[i][j] = min(dp[i][k]+dp[k+1][j]+a[i].pl*a[k].pr*a[j].pr);// 在区间i到j上找一个k使得dp[i][k]+dp[k+1][j]这两部分的和在加上最后的// a[i].pl*a[k].pr*p[i].pr的最小值;// 能有这样的状态关键是; P =A[1] * A[2] * …. * A[K] // 和 Q= A[K+1] * A[K+2] * …. * A[N] 这两部分的结果是不相互影响的，// 因此只需要分别让 P 和 Q 最优方法计算，最优子结构！// 边界时dp[i][i]=0;//// 这是一道痕经典的区间dp的题目，学会了用这种方法考虑问题// 最后，，还有就是递归打印，实在是十分奇妙。// 还是挺高兴的呢。哎，继续练吧。。。。#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cfloat>#include <climits>#include <cmath>#include <complex>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <vector>#define ceil(a,b) (((a)+(b)-1)/(b))#define endl '\n'#define gcd __gcd#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))#define popCount __builtin_popcountlltypedef long long ll;using namespace std;const int MOD = 1000000007;const long double PI = acos(-1.L);template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }template<class T> inline T lowBit(const T& x) { return x&-x; }template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }const int maxn = 1008;struct node {int pl;int pr;}a[maxn];ll d[maxn][maxn];int path[maxn][maxn];int n;const ll inf = 0x4f4f4f4f4f4f4f4f;ll dp(int i,int j){if (d[i][j]!=inf)return d[i][j];ll& ans = d[i][j];for (int k=i;k<j;k++){ll res = dp(i,k)+dp(k+1,j)+(ll)a[i].pl*a[k].pr*a[j].pr;if (ans>res){ans = res;path[i][j] = k;}}return ans;}void print(int i,int j){if (i==j){printf("A%d",i+1);return;}printf("(");for (int k=i;k<j;k++)if (path[i][j]==k){print(i,k);printf(" x ");print(k+1,j);break;}printf(")");//printf(")");}int main() {int kase = 0;//freopen("G:\\Code\\1.txt","r",stdin);while(scanf("%d",&n)!=EOF&&n){for (int i=0;i<n;i++)scanf("%d%d",&a[i].pl,&a[i].pr);//memset(d,0x4f,sizeof(d));//memset(path,0,sizeof(path));for (int i=0;i<n;i++)for (int j=0;j<n;j++){path[i][j]=0;d[i][j]=inf;}for (int i=0;i<n;i++)d[i][i]=0;dp(0,n-1);//printf("%lld\n",dp(0,n-1));printf("Case %d: ",++kase);print(0,n-1);puts("");}return 0;}
任何业绩的质变都来自于量变的积累。
