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Given a matrix with n rows and m columns ( n+m is an odd number ), at first , you begin with the number at top-left corner (1,1) and you want to go to the number at bottom-right corner (n,m). And you must go right or go down every steps. Let the numbers you go through become an array a1,a2,...,a2k. The cost isa1∗a2+a3∗a4+...+a2k−1∗a2k. What is the minimum of the cost?
Several test cases(about 5) For each cases, first come 2 integers, n,m(1≤n≤1000,1≤m≤1000) N+m is an odd number. Then follows n lines with m numbers ai,j(1≤ai≤100)
For each cases, please output an integer in a line as the answer.
2 3 1 2 3 2 2 1 2 3 2 2 1 1 2 4
4 8
令dp[i][j]表示当前走到第i,j个位置的最小贡献，初始化做好了，然后根据i+j是奇数偶数的情况分别计算dp即可，最后要用long long。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 1006 23 #define inf 1<<29 24 ll n,m; 25 ll mp[N][N]; 26 ll dp[N][N]; 27 int main() 28 { 29 while(scanf("%I64d%I64d",&n,&m)==2){ 30 memset(dp,0,sizeof(dp)); 31 for(ll i=1;i<=n;i++){ 32 for(ll j=1;j<=m;j++){ 33 scanf("%I64d",&mp[i][j]); 34 } 35 } 36 for(ll i=0;i<=n+1;i++){ 37 dp[i][0]=inf; 38 mp[i][0]=inf; 39 } 40 for(ll i=0;i<=m+1;i++){ 41 dp[0][i]=inf; 42 mp[0][i]=inf; 43 } 44 45 dp[1][1]=mp[1][1]; 46 dp[1][2]=mp[1][1]*mp[1][2]; 47 dp[2][1]=mp[1][1]*mp[2][1]; 48 for(ll i=1;i<=n;i++){ 49 for(ll j=1;j<=m;j++){ 50 if(i==1 && j==1) continue; 51 if(i==1 && j==2) continue; 52 if(i==2 && j==1) continue; 53 if((i+j)%2==0){ 54 dp[i][j]=min(dp[i-1][j],dp[i][j-1]); 55 //printf("i , j %d %d %d\n",i,j,dp[i][j]); 56 } 57 else{ 58 dp[i][j]=min(dp[i-1][j]+mp[i-1][j]*mp[i][j],dp[i][j-1]+mp[i][j-1]*mp[i][j]); 59 //printf("i , j %d %d %d\n",i,j,dp[i][j]); 60 } 61 62 } 63 } 64 65 printf("%I64d\n",dp[n][m]); 66 } 67 return 0; 68 }
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原文地址：http://www.cnblogs.com/UniqueColor/p/4984789.html
