hdu5569 BestCoder Round #63 (div.2)

题意：

给你一个矩阵，要求从左上角走到右下角，走个的费用：a[1]*a[2] + a[3]*a[4] + ......+ a[2n-1]*a[2n]

思路：

果然不机智，自己把自己套路了

对于每个奇数点，如下图的有下角的点它便可由3个值为2的点到达，具体画图便知。

所以我们可以用类似dp的方法，找出每个点的奇数点最优解，注意下边界即可

1 1 1 2

1 1 2 1

1 2 1 1

 

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#include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <algorithm> const int inf = 0x3f3f3f3f; using namespace std; typedef long long ll; ll a[1005][1005];   int main() {     int n,m;     while(scanf("%d%d",&n,&m) != EOF)     {         for(int i = 1; i <= n; i++)             for(int j = 1; j <= m; j++)             {                 scanf("%I64d",&a[i][j]);             }         a[1][2] *= a[1][1];         a[2][1] *= a[1][1];         for(int i = 1; i <= n; i++)             for(int j = 1; j <= m; j++)             {                 if((i+j)%2)                 {                     int t1,t2,t3;                     t1 = t2 = t3 = inf;                     if(j > 2)                         t1 = a[i][j-2]+a[i][j]*a[i][j-1];                     if(i > 1 && j > 1)                         t2 = a[i-1][j-1]+min(a[i][j]*a[i-1][j],a[i][j]*a[i][j-1]);                     if(i > 2)                         t3 = a[i-2][j]+a[i-1][j]*a[i][j];                     t1 = min(t1,t2);                     t1 = min(t1,t3);                     if(t1 >= inf)                         continue;                     a[i][j] = t1;                 }             }         printf("%I64d\n",a[n][m]);     }     return 0; }