POJ2152 Fire 【树形dp】

题目链接

题解

经典老题，还真暴力

\(n \le 1000\)，所以可以\(O(n^2)\)做

所以可以枚举每个点依附于哪一个点

设\(f[u]\)表示以\(u\)为根的子树的最小代价

\(g[u][v]\)表示\(u\)依附于\(v\)时以\(u\)为根的子树的最小代价

显然

\[f[u] = min\{ g[u][v] \}
\]

\[g[u][v] = cost[v] + \sum\limits_{(u,to) \in edge} min(g[to][v] - cost[v],f[to]) \quad [dis(u,v) \le D(u)]
\]

\(dis(u,v)\)要直接暴力\(O(n^2)\)预处理

上倍增直接\(T\)掉。。。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define cls(s) memset(s,0,sizeof(s))

using namespace std;

const int maxn = 1005,maxm = 100005,INF = 1000000000;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int h[maxn],ne;

struct EDGE{int to,nxt,w;}ed[maxn << 1];

inline void build(int u,int v,int w){

	ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;

	ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;

}

int fa[maxn],f[maxn],g[maxn][maxn],cost[maxn],d[maxn],dis[maxn][maxn],n,rt;

void DFS(int u,int D,int F){

	dis[rt][u] = D;

	Redge(u) if ((to = ed[k].to) != F)

		DFS(to,D + ed[k].w,u);

}

void dfs(int u){

	f[u] = INF;

	REP(i,n)

		if (dis[u][i] <= d[u]) g[u][i] = cost[i];

		else g[u][i] = INF;

	Redge(u) if ((to = ed[k].to) != fa[u]){

		fa[to] = u; dfs(to);

		REP(i,n) if (g[u][i] != INF)

			g[u][i] += min(g[to][i] - cost[i],f[to]);

	}

	REP(i,n) f[u] = min(f[u],g[u][i]);

}

int main(){

	int T = read();

	while (T--){

		n = read(); ne = 2; cls(h);

		REP(i,n) cost[i] = read();

		REP(i,n) d[i] = read();

		int a,b,w;

		for (int i = 1; i < n; i++){

			a = read(); b = read(); w = read();

			build(a,b,w);

		}

		REP(i,n) rt = i,DFS(i,0,0);

		dfs(1);

		printf("%d\n",f[1]);

	}

	return 0;

}

巴特西

POJ2152 Fire 【树形dp】

题目链接

题解

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