http://codevs.cn/submission/2367697/
loli蜜汁(面向高一)树形dp是这道题的改编。
改编后的题目中每个展览厅的有多个不同的画,偷画的时间和画的价值也不同,求最大价值。
需要在叶节点上做01背包。
但codevs上的这道题就简单多了,直接改了改01背包交上去A了。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct nodeTreeDP {
struct node {int nxt, to;} E[100003];
int cnt, val[103], point[50003], f[103][603], m, tot;
nodeTreeDP() {
cnt = 0; tot = 1;
memset(E, 0, sizeof(E));
memset(f, 0, sizeof(f));
memset(point, 0, sizeof(point));
}
void ins(int u, int v) {E[++cnt] = (node) {point[u], v}; point[u] = cnt;}
void dfsread(int from) {
int t, x;
scanf("%d%d", &t, &x);
++tot;
ins(from, tot);
val[tot] = t << 1;
if (x) {
int w, c; w = 1; c = 5;
for(int i = 1; i <= x; ++i) {
for(int j = 600; j >= 0; --j)
if (j - (t << 1) - c >= 0)
f[tot][j] = max(f[tot][j], f[tot][j - c] + w);
}
} else {
int now = tot;
dfsread(now);
dfsread(now);
}
}
void readin() {
int t, x;
scanf("%d%d", &t, &x);
val[1] = t << 1;
if (!x) {
dfsread(1);
dfsread(1);
} else {
int w, c; w = 1; c = 5;
for(int i = 1; i <= x; ++i) {
for(int j = 600; j >= 0; --j)
if (j - (t << 1) - c >= 0)
f[1][j] = max(f[1][j], f[1][j - c] + w);
}
}
}
void TreeDP(int x) {
for(int i = point[x]; i; i = E[i].nxt)
TreeDP(E[i].to);
if (!point[x]) return;
int left = 0, right = 0;
for(int i = point[x]; i; i = E[i].nxt) {
if (!left) left = E[i].to;
else right = E[i].to;
}
for(int top = val[x]; top <= 600; ++top) {
int res = top - val[x];
for(int leftnum = 0; leftnum <= res; ++leftnum) {
int rightnum = res - leftnum;
f[x][top] = max(f[x][top], f[left][leftnum] + f[right][rightnum]);
}
}
}
void ansit() {
TreeDP(1);
printf("%d
", f[1][m]);
}
} *T;
int main() {
T = new nodeTreeDP;
scanf("%d", &T->m);
T->readin();
T->ansit();
return 0;
}