[poj 2152]Fire

Posted 2020-10-11

题目大意：给你一棵树，边上权值表示距离，一个点上可以建消防站，花费d[i]一个节点要么建消防站，要么周围dis[i]距离内必须有一个消防站，求覆盖全图的最小花费.

 

设f[i][j]为点i被火炉j控制的最优解.b[i]为点i最优方案.

\({b_i} = \min \{ {f_{i,j}}|j \in {s_i}\} \),{s_i}是i的儿子集合.

\({f_{i,j}} = {w_j} + \sum\limits_{k \in {s_i}} {\min \{ {b_k},{f_{k,j}} - {w_j}\} } \)

 

代码如下:

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 #define N 1010
 5 #define inf 0x2fffffff
 6 #define __min(a,b) (((a)<(b))?(a):(b))
 7 int T,n,tot;int head[N],next[N << 1],to[N << 1],val[N << 1],cost[N],md[N],co[N],f[N][N],best[N],dis[N];
 8 void addedge(int u,int v,int w){next[tot] = head[u];to[tot] = v;val[tot] = w;head[u] = tot++;}
 9 std::queue<int> q;int dist(int s)
10 {
11     memset(dis,-1,sizeof(dis));dis[s] = 0;q.push(s);
12     while(!q.empty())
13     {
14         int v = q.front();q.pop();for(int i = head[v];~i;i = next[i])
15         {
16             if(~dis[to[i]]) continue;
17             dis[to[i]] = dis[v] + val[i];q.push(to[i]);
18         }
19     }
20 }
21 void dfs(int u,int fa)
22 {
23     for(int i = head[u];~i;i = next[i]) if(to[i] != fa) dfs(to[i],u);dist(u);
24     best[u] = inf;for(int i = 0;i < n;i++)
25     {
26         if(dis[i] > md[u]) f[u][i] = inf;
27         else
28         {
29             f[u][i] = co[i];for(int j = head[u];~j;j = next[j]) if(to[j] != fa) f[u][i] += __min(best[to[j]],f[to[j]][i] - co[i]);
30             best[u] = __min(best[u],f[u][i]);
31         }
32     }
33 }
34 int main()
35 {
36     scanf("%d",&T);while(T--)
37     {
38         memset(head,-1,sizeof(head));scanf("%d\n",&n);for(int i = 0;i < n;i++)scanf("%d",&co[i]);for(int i = 0;i < n;i++)scanf("%d",&md[i]);
39         tot = 0;for(int i = 1,u,v,w;i < n;i++){scanf("%d %d %d",&u,&v,&w);addedge(u - 1,v - 1,w);addedge(v - 1,u - 1,w);}
40         dfs(0,-1);printf("%d\n",best[0]);
41     }
42     return 0;
43 }