但这一题不能贪心，只能采用dp的方法。形式上却也差不多，都是对子树的讨论。令f[i][j]为以i为根的子树，能向子树外拓展i个节点最少需要炸毁几个城市。G[i][j]为以i为根的子树，子树内有节点未被炸毁，且距离根为j最少需要炸毁几个城市。
转移方程:
不炸毁u点
f[u][j]=f[v][j+1]+min(f[k][0 j+1],G[k][0 j]);
G[u][0]=f[u][0];
G[u][j]=G[v][j?1]+min(f[k][0 j?1],G[k][0 j?1]);
炸毁u点
f[u][w[u]]=1+min(f[v][0 w[u]+1],G[v][w[u]]);
这里需要对f[][]和G[][],求前缀最小值。
节点比较多，避免爆栈不能直接用递归的方法做。
时间复杂度O（n?w）
Bombing plan
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 364 Accepted Submission(s): 86
Problem Description
Kingdom Y is in the war with kingdom X. Kingdom X consists of N cities,there are N-1 bidirectional roads which are all 1 long ，each of them connect a pair of cities,the N cities are all connect by the N-1 bidirectional.People can travel through the roads.
Now kingdom Y is going to bomb kingdom X. Every city of kingdom X has its own value W. If city i was to be bombed, then all the cities that lie within the distance W(i) from city i would be destroyed as well. The king of kingdom Y wants to know the minimum
bombing time that can destroy all the cities in kingdom X. Could you help him?
Input
There are multiple test cases. Please process till EOF.
In each test case:
First line: an integer n(n<=10^5) indicating the number of city
Second line:contain n numbers w[i](0<=w[i]<=100) ,indicating that the value of city[i],
Next n - 1 lines: each contains two numbers ui and vi, (1 ≤ ui,vi<=n), indicates that there’s one road connecting city ui and vi.
Output
For each case,output one number, denotes the minimum number of bombing times.
Sample Input
5 1 1 1 1 1 1 2 2 3 3 4 4 5
Sample Output
2
Author
FZUACM
Source
2015 Multi-University Training Contest 1
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define prt(k) cout<<#k" = "<<k<<endl;
const int inf = 0x3f3f3f3f;
const int N = 100007;
const int W = 107;
int f[N][107], g[N][107];
int father[N];
int n;
int w[N];
int F[N][107], G[N][107];
int maxw;
struct edge
{
int v, next;
}e[N<<1];
int head[N], mm;
void add(int u, int v)
{
e[mm].v = v;
e[mm].next = head[u];
head[u] = mm++;
}
struct P
{
int d, id;
P() {}
P(int _id, int _d=0) { d=_d, id=_id; }
bool operator < (P b) const
{
return d > b.d;
}
}p[N];
bool vis[N];
void bfs()
{
queue<P> q;
q.push(P(1,0));
memset(vis,0,sizeof vis);
vis[1] = true;
father[1] = 1;
while (!q.empty()) {
P u = q.front(); q.pop();
int now = u.id;
p[now].id = now;
p[now].d = u.d + 1;
for (int i=head[now];~i;i=e[i].next) {
int v = e[i].v;
if (!vis[v]) {
q.push(P(v, u.d + 1));
vis[v] = true;
father[v] = now;
}
}
}
}
void gao(int u, int fa)
{
assert(!vis[u]);
vis[u] = true;
bool flag = false;
for (int i=head[u];~i;i=e[i].next) {
int v = e[i].v;
if (v==fa) continue;
flag = true;
assert(vis[v]);
}
if (!flag) {
g[u][0] = 0;
f[u][w[u]] = 1;
}
for (int j=0;j<=maxw;j++) {
ll sum=0;
for (int i=head[u];~i;i=e[i].next) {
int v = e[i].v;
if (v==fa) continue;
int t = F[v][j+1];
if (j > 0)
t = min(t, G[v][j-1]);
sum += t;
if (sum >= inf) { sum = inf; break; }
}
if (sum < inf ) for (int i=head[u];~i;i=e[i].next) {
int v = e[i].v;
if (v==fa) continue;
int t = F[v][j+1];
if (j > 0) t = min(t, G[v][j-1]);
if (0<=sum && sum < inf) f[u][j] = min(f[u][j], (int)(f[v][j+1] + sum - t ) ) ;
}
sum = 0;
int t;
for (int i=head[u];~i;i=e[i].next) {
int v = e[i].v;
if (v==fa) continue;
int t = F[v][j];
if (j > 0) t = min(t, G[v][j-1]);
sum += t;
if (sum >= inf) { sum = inf; break; }
}
if (j>0 && sum < inf) for (int i=head[u];~i;i=e[i].next) {
int v = e[i].v;
if (v==fa) continue;
int t = F[v][j];
if (j > 0) t = min(t, G[v][j-1]);
if (0<=sum && sum < inf) g[u][j]=min(g[u][j], (int)(g[v][j-1] + sum - t ) );
}
}
ll sum = 0;
for (int i=head[u];~i;i=e[i].next) {
int v = e[i].v;
if (v==fa) continue;
sum += min(F[v][w[u]+1], w[u]>0 ? G[v][w[u]-1] : inf);
}
if (0<=sum && sum<inf) f[u][w[u]]=min(f[u][w[u]], int(1 + sum ));
F[u][0] = f[u][0];
G[u][0] = g[u][0];
for (int j=1;j<=maxw+1;j++) {
F[u][j] = min(F[u][j-1], f[u][j]);
G[u][j] = min(G[u][j-1], g[u][j]);
}
}
int main()
{
while (scanf("%d", &n)==1) {
maxw = 0;
for (int i=1;i<=n;i++) scanf("%d", w+i), maxw=max(maxw, w[i]);
mm=0; memset(head,-1,sizeof head);
for (int i=1;i<=n-1;i++)
{
int u, v; scanf("%d%d", &u, &v);
add(u, v);
add(v, u);
}
memset(f, 63, sizeof f);
memset(g, 63, sizeof g);
bfs();
sort(p+1, p+n+1);
memset(vis, 0, sizeof vis);
for (int i=1;i<=n;i++) {
gao(p[i].id, father[p[i].id]);
}
int ans = n;
for (int i=0;i<=maxw;i++)
ans = min(ans, f[1][i]);
printf("%d\n", ans);
}
}
/**
3
1 1 1
1 2
2 3
7
1 1 1 1 1 1 1
1 2
2 3
3 4
4 5
5 6
6 7
4
0 1 0 1
1 2
1 3
1 4
4
0 2 0 0
1 2
1 3
1 4
*/
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