[POJ1155]TELE

試題描述

A TV-network plans to broadcast an important football match. Their network of transmitters and users can be represented as a tree. The root of the tree is a transmitter that emits the football match, the leaves of the tree are the potential users and other vertices in the tree are relays (transmitters).?
The price of transmission of a signal from one transmitter to another or to the user is given. A price of the entire broadcast is the sum of prices of all individual signal transmissions.?
Every user is ready to pay a certain amount of money to watch the match and the TV-network then decides whether or not to provide the user with the signal.?
Write a program that will find the maximal number of users able to watch the match so that the TV-network's doesn't lose money from broadcasting the match.

輸入

The first line of the input file contains two integers N and M, 2 <= N <= 3000, 1 <= M <= N-1, the number of vertices in the tree and the number of potential users.?
The root of the tree is marked with the number 1, while other transmitters are numbered 2 to N-M and potential users are numbered N-M+1 to N.?
The following N-M lines contain data about the transmitters in the following form:?
K A1 C1 A2 C2 ... AK CK?
Means that a transmitter transmits the signal to K transmitters or users, every one of them described by the pair of numbers A and C, the transmitter or user's number and the cost of transmitting the signal to them.?
The last line contains the data about users, containing M integers representing respectively the price every one of them is willing to pay to watch the match.

輸出

The first and the only line of the output file should contain the maximal number of users described in the above text.

輸入示例

9 6 3 2 2 3 2 9 3 2 4 2 5 2 3 6 2 7 2 8 2 4 3 3 3 1 1

輸出示例

5

數據規模及約定

見“輸入”；另：過程中不會有超過 int 的值。

題解

樹形 dp（樹上背包）。

設 f(i, j) 表示子樹 i 中選擇了 j 個葉子的最大獲利（若為負則 -f(i, j) 為最小虧損）。那么答案就是最大的 j，滿足 f(i, j) 非負。

考慮子樹 u，兒子上的信息肯定是最有子結構，所以先算出所有的 f(son, j)，然后分別將一個個子樹的信息加入 f(i, j)（f(u, i+j) = max{ f(u, i) + f(son, j) - dist(i, son) | j > 0 , f(u, i) + f(son, j) | j = 0 }）。

可以證明總轉移數是 O(n²) 級別的，詳見這里。

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <algorithm> using namespace std;const int BufferSize = 1 << 16; char buffer[BufferSize], *Head, *Tail; inline char Getchar() {if(Head == Tail) {int l = fread(buffer, 1, BufferSize, stdin);Tail = (Head = buffer) + l;}return *Head++; } int read() {int x = 0, f = 1; char c = Getchar();while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }return x * f; }#define maxn 3010 #define oo 2147483647int n, usr, m, head[maxn], nxt[maxn], to[maxn], dist[maxn], pay[maxn];void AddEdge(int a, int b, int c) {to[++m] = b; dist[m] = c; nxt[m] = head[a]; head[a] = m;return ; }int f[maxn][maxn], clea[maxn]; void dp(int u) {if(u > n - usr) {clea[u] = 1;f[u][0] = 0; f[u][1] = pay[u];return ;}f[u][0] = 0;for(int e = head[u]; e; e = nxt[e]) {dp(to[e]);for(int i = clea[u]; i >= 0; i--) if(f[u][i] < oo)for(int j = 0; j <= clea[to[e]]; j++) if(f[to[e]][j] < oo)f[u][i+j] = max(f[u][i+j], f[u][i] + f[to[e]][j] - (j ? dist[e] : 0));clea[u] += clea[to[e]];}return ; }int main() {n = read(); usr = read();for(int i = 1; i <= n - usr; i++) {int k = read();while(k--) {int u = read(), c = read();AddEdge(i, u, c);}}for(int i = n - usr + 1; i <= n; i++) pay[i] = read();for(int i = 1; i <= n; i++)for(int j = 0; j <= n; j++) f[i][j] = -oo;dp(1);for(int j = clea[1]; j; j--) if(f[1][j] >= 0) return printf("%d\n", j), 0;puts("0");return 0; }

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轉載于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/7309030.html

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