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DP过~~(矩阵乘法)#include <assert.h> #include <ctype.h> #include <errno.h> #include <float.h> #include <fstream> #include <iomanip> #include <iostream> #include <limits> #include <deque> #include <locale> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <time.h> #include <wchar.h> #include <wctype.h> #include <algorithm> #include <bitset> #include <map> #include <iomanip> #include <ios> #include <iostream> #include <vector> #include <cwchar> #include <cwctype> #define mp make_pair #define fs first #define se second #define memset(a,t) memset(a,t,sizeof(a)) #define all(v) v.begin(),v.end() #define eprintf(...) fprintf(stderr, __VA_ARGS__),fflush(stderr) #define MN 0LL #define MX 20000000000000005 using namespace std; long long dp[105][105]; int main(){ // freopen("input.txt","r",stdin); // freopen("output.txt","w",stdout); ios_base::sync_with_stdio(false); long long n,i,j,k,s; for(i=1;i<105;i++) for(j=1;j<105;j++) dp[i][j]=MX; long long m[105]; cin>>n; for(i=1;i<=n;i++) cin>>m[i]; for(i=1;i<=n;i++) dp[i][i]=0; for(s=1;s<=n-1;s++){ for(i=1;i<=n-s;i++){ j=i+s; for(k=i;k<j;k++){ dp[i][j]=min(dp[i][j],dp[k+1][j]+dp[i][k]+m[i-1]*m[k]*m[j]); } } } // for(i=1;i<=n;i++){ // for(j=1;j<=n;j++) cout<<dp[i][j]<<" "; // cout<<endl; // } cout<<dp[2][n]<<endl; return 0; #ifdef home eprintf("time = %d ms\n", (int)(clock() * 1000. / CLOCKS_PER_SEC)); #endif } Followed by: Post your reply here: |
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