sgu104：Little Shop of Flowers

设f[i][j]表示前i个花瓶装下j朵花的最大价值，则：
如果第i个花瓶不装，f[i][j]=f[i-1][j];
如果第i个花瓶装，f[i][j]=f[i-1][j-1]+val[j][i];
因此f[i][j]=max{f[i-1][j],f[i-1][j-1]+val[j][i]};
max{f[i][F]}(F<=i<=V)即为第一问的答案

至于第二问，倒推回去即可

比较水的DP题

代码如下：

#include <stdio.h>#include <string.h>int f[105][105] = {0};int F, V;int val[105][105] = {0};int main(){  int i, j;  int path[105] = {0};  int ans = -2147483647, flag = 0;  scanf("%d%d", &F, &V);  for(i = 1; i <= F; ++i)    for(j = 1; j <= V; ++j)      scanf("%d", &val[i][j]);    memset(f, -1e6, sizeof(f));  for(i = 0; i <= V; ++i) f[i][0] = 0;  for(i = 1; i <= V; ++i)    for(j = 1; j <= F && j <= i; ++j)    {      f[i][j] = f[i - 1][j];  if(f[i][j] < f[i - 1][j - 1] + val[j][i])    f[i][j] = f[i - 1][j - 1] + val[j][i];    }    for(i = F; i <= V; ++i)    if(f[i][F] > ans) ans = f[i][F], flag = i;  printf("%d\n", ans);    j = flag;  for(i = F; i >= 1; --i)  {    while(1)    {  if(f[j - 1][i] == f[j][i]) j--;      else if(f[j - 1][i - 1] + val[i][j] == f[j][i])       {        path[i] = j--;break;      }    }  }  for(i = 1; i <= F; ++i)    printf("%d ", path[i]);  return 0;}