非常经典的皇宫守护问题,用树形DP即可求解。对于每个点的覆盖可以分三种情况讨论:
- 在该点设守卫
- 在子节点设守卫
- 在父节点设守卫
分别用f[i,1],f[i,2],f[i,3]表示以上三种情况覆盖以该节点为根的子树所需的最少费用。则显然可以推出状态转移方程:
- f[i,1]=∑min(f[j,1],f[j,2],f[j,3])+v[i](j为i的子节点);
- f[i,2]=∑min(f[j,1],f[j,2])+mind(j为i的子节点;mind为所有子节点中f[j,1]-f[j,2]最小的,若存在f[j,1]<=f[j,2]则mind为0);对于叶节点,f[i,2]=v[i]。
- f[i,3]=∑f[j,2](j为i的子节点);
程序实现时用一个队列维护当前待处理的已经处理完全部叶节点的节点,最后加入队列中的节点就是根节点。
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