UVA10672 POJ1909 ZOJ2374 Marbles on a treeBFS

Posted 2022-04-28 海岛Blog

Marbles on a tree
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 1890 Accepted: 885

Description

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?

Input

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.

Output

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

Sample Output

7
14
20

Source

Waterloo local 2004.06.12

问题链接：UVA10672 POJ1909 ZOJ2374 Marbles on a tree
问题简述：（略）
问题分析：用BFS来解决，不解释。这个题分类上属于用拓扑排序来解决。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10672 POJ1909 ZOJ2374 Marbles on a tree */

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cstdlib>

using namespace std;

const int N = 10000;
int degree[N], marbles[N], p[N];

int main()

    memset(degree, 0, sizeof(degree));
    memset(marbles, 0, sizeof(marbles));
    memset(p, 0, sizeof(p));

    int n, v, m, d;
    while (~scanf("%d", &n) && n) 
        for (int i = 0; i < n; i++) 
            scanf("%d%d%d", &v, &m, &d);
            degree[v] = d;
            marbles[v] = m;
            for (int j = 0; j < d; j++) 
                int child;
                scanf("%d", &child);
                p[child] = v;
            
        

        queue<int> q;
        for (int i = 1; i <= n; i++)
            if (degree[i] == 0) q.push(i);

        int ans = 0;
        while (!q.empty()) 
            int u = q.front();
            q.pop();
            int parent = p[u];
            if (parent == 0) break;
            if (marbles[u] != 1) 
                marbles[parent] += marbles[u] - 1;
                ans += abs(marbles[u] - 1);
                marbles[u] = 1;
            

            if (--degree[parent] == 0) q.push(parent);
        

        printf("%d\\n", ans);
    

    return 0;