2015
04-14

# Replica Placement

Topsky wants to build a content delivery network. It contains an origin server and some mirror servers as shown in figure 1. Original data is put at the origin server (root in figure 1) while replicas are put at some of the mirror servers (node 2 and 5 in figure 1). When a node issues a request of data, it will try to find its destination in following steps.
1.  Check if there is a replica at its own place. If yes, the request meets. Otherwise do step 2.
2.  Forward the request to its parent, and let its parent do step 1.

The cost of meeting the request C (v) is defined as the sum of weight of the edges along the road. If C (v) is not greater than an upper bound Q (v), then the retrieval cost is satisfied. Topsky further assumes a nonnegative cost S (v) which means the cost of storing data at node v. Note that the origin server is special, the cost of storing data at origin server is 0. Now Topsky wants to find the way of replicas placement such the retrieval cost of all nodes are satisfied while the total storage cost is minimal.

The first line contains a single integer T (T <= 20), indicating the number of test cases.
Each case begins with three integers N (1<= N <= 1000) indicates the number of servers.
Then N lines follow. Each line contains four integers Fv (0 <= Fv <= N), Qv, Sv and Wv (0 <= Qv, Sv, Wv <= 105). In line i (1 <= i <= N), Fv is the father of node i (if Fv is 0, it means node i is the origin server, Qv is -1, Sv is 0 and Wv is 0), Qv is upper bound of retrieval cost, Sv is storing cost of node i and Wv is weight of this edge between node i and node Fv.

The first line contains a single integer T (T <= 20), indicating the number of test cases.
Each case begins with three integers N (1<= N <= 1000) indicates the number of servers.
Then N lines follow. Each line contains four integers Fv (0 <= Fv <= N), Qv, Sv and Wv (0 <= Qv, Sv, Wv <= 105). In line i (1 <= i <= N), Fv is the father of node i (if Fv is 0, it means node i is the origin server, Qv is -1, Sv is 0 and Wv is 0), Qv is upper bound of retrieval cost, Sv is storing cost of node i and Wv is weight of this edge between node i and node Fv.

1
3
0 -1 0 0
1 1 1 1
2 1 1 1

1

/*
* =====================================================================================
*
*       Filename:  dp.cpp
*
*    Description:  Problem A on ChengDu
*
*        Version:  1.0
*        Created:  2011/6/23 21:11:48
*       Revision:  none
*       Compiler:  gcc
*
*         Author:  ronaflx
*        Company:  hit-acm-group
*
* =====================================================================================
*/
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <climits>
#include <algorithm>
#include <functional>
#include <numeric>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <list>
#include <string>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <stdexcept>
#include <utility>
#include <cassert>
#include <complex>
using namespace std;
#define DEBUG(x) cout << #x << " " << x << endl
#define LEFT(i) ((i) << 1)
#define RIGHT(i) (((i) << 1) | 1)
#define MID(i) ((l[i] + r[i]) >> 1)
#define CC(i, v) memset(i, v, sizeof(i))
#define REP(i, l, n) for(int i = l;i < int(n);++i)
#define FOREACH(con, i) for(__typeof(con.begin()) i = con.begin();i != con.end();++i)
const int INF = 100000000;
const int N = 1008;
int dp[N][N];
int s[N], q[N], f[N], n, w[N];
bool isLeaf(int u)
{
}

void dfs(int u)
{
dfs(*i);
int now = 0, v;
if(isLeaf(u))
{
for(v = u; v != -1;now += w[v], v = f[v])
{
if(v == u || now > q[u])
dp[u][v] = s[u];
else
dp[u][v] = 0;
}
}
else
{
for(v = u;v != -1;now += w[v], v = f[v])
{
int tot = 0;
tot += min(dp[*j][v], dp[*j][*j]);
if(u == v || now > q[u])
dp[u][v] = tot + s[u];
else
dp[u][v] = min(tot, dp[u][u]);
}
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
int root = -1;
for(int i = 0;i < n;i++) adj[i].clear();
for(int i = 0;i < n;i++)
{
scanf("%d %d %d %d", &f[i], &q[i], &s[i], &w[i]);
f[i]--;
if(f[i] == -1)
{
root = i;
q[i] = 0;
}
}
dfs(root);
printf("%d\n", dp[root][root]);
}
return 0;
}

1. 我没看懂题目
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
第二个是7 0 6 -1 1 -6 7输出是14 7 7
不知道题目例子是怎么得出来的