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
#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 ",a[n][m]); } return 0; }