[UVA] 10672 - Marbles on a tree
Problem E: Marbles on a tree
n boxes are placed on the vertices of a rooted tree, which
are numbered from 1 to n,
1 ≤ n ≤ 10000. Each box is either
empty or contains a number of marbles; the total number of marbles is
n.
The task is to move the marbles such that each box contains exactly one marble. This is to be accomplished be a sequence of moves; each move consists of moving one marble to a box at an adjacent vertex. What is the minimum number of moves required to achieve the goal?
The input contains a number of cases. Each case starts with the number n followed by n lines. Each line contains at least three numbers which are: v the number of a vertex, followed by the number of marbles originally placed at vertex v followed by a number d which is the number of children of v, followed by d numbers giving the identities of the children of v.
The input is terminated by a case where n = 0 and this case should not be processed.
For each case in the input, output the smallest number of moves of marbles resulting in one marble at each vertex of the tree.
Sample input
9 1 2 3 2 3 4 2 1 0 3 0 2 5 6 4 1 3 7 8 9 5 3 0 6 0 0 7 0 0 8 2 0 9 0 0 9 1 0 3 2 3 4 2 0 0 3 0 2 5 6 4 9 3 7 8 9 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 9 1 0 3 2 3 4 2 9 0 3 0 2 5 6 4 0 3 7 8 9 5 0 0 6 0 0 7 0 0 8 0 0 9 0 0 0
Output for sample input
7 14 20
P. Rudnicki
如同中山大學的呂為萱所寫的描述去解題，她的構想其實就是對於這個點有多少彈珠會經過，這個的計算只需要看成子樹總共缺多少彈珠，又或者是多多少彈珠，這兩個值得大小就是其經過的次數。
#include <stdio.h>
#include <stdlib.h>
#include <vector>
using namespace std;
struct ND {
int totm;
vector<int> son;
};
ND T[10001];
long long ans = 0;
int used[10001];
int sol(int nd) {
used[nd] = 1;
int cnt = 0;
for(vector<int>::iterator it = T[nd].son.begin();
it != T[nd].son.end(); it++) {
if(used[*it] == 0) {
cnt += sol(*it);
T[nd].totm += T[*it].totm;
}
}
ans += abs(T[nd].totm-cnt-1);
return cnt+1;
}
int main() {
int n, v, m, s, x;
int i, j;
while(scanf("%d", &n) == 1 && n) {
for(i = 1; i <= n; i++)
T[i].son.clear(), used[i] = 0;
for(i = 1; i <= n; i++) {
scanf("%d %d %d", &v, &m, &s);
T[v].totm = m;
while(s--) {
scanf("%d", &x);
T[v].son.push_back(x);
T[x].son.push_back(v);
}
}
ans = 0;
sol(1);
printf("%lld\n", ans);
}
return 0;
}
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