POJ-1655（DFS）

最新推荐文章于 2019-08-13 21:52:00 发布

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最新推荐文章于 2019-08-13 21:52:00 发布

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分类专栏：每天A一道题 DFS

本文链接：https://blog.youkuaiyun.com/uuuououlcz/article/details/37546377

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本文探讨了一种基于图遍历的算法优化方案，通过改进后序遍历减少内存占用，并对比了不同实现方式对运行时间的影响。

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#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int n, index, balance;
vector<int> neighbour[20001];
bool vis[20001];
int subtree[20001];

void postTraverse(int i)
{
    vis[i] = true;
    subtree[i] = 1;
    const vector<int>& v = neighbour[i];
    int tmp = 0, j = 0, s = v.size(), k;
    for(; j < s; ++j){
        k = v[j];
        if(!vis[k]){
            postTraverse(k);
            subtree[i] += subtree[k];
            tmp = max(tmp, subtree[k]);
        }
    }
    tmp = max(tmp, n - subtree[i]);
    if(tmp < balance || tmp == balance && i < index){
        balance = tmp;
        index = i;
    }
}

int main()
{
    int test, i, x, y;
    for(scanf("%d", &test); test--; ){
        scanf("%d", &n);
    //initialize
        for(i = 1; i <= n; ++i){
            neighbour[i].clear();
            vis[i] = false;
        }
        balance = 20002;
    //input edges
        for(i = 1; i < n; ++i){
            scanf("%d%d", &x, &y);
            neighbour[x].push_back(y);
            neighbour[y].push_back(x);
        }
    //find a leave to be an root
        for(i = 1; i <= n; ++i){
            if(neighbour[i].size() == 1) break;
        }
    //post traverse and find each subtree's nodes count, then solve the problem
        postTraverse(i);
        printf("%d %d\n", index, balance);
    }
    return 0;
}

实际上，我们并不需要找出一个leave，而且上面的vis可以用带parent的DFS省去，subtree[]可以用postTraverse返回值省去，这样我们还可以省去20001*(sizeof(int) + sizeof(bool)) ~100K的空间：

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int n, index, balance;
vector<int> neighbour[20001];

int postTraverse(int i, int pre)
{
    const vector<int>& v = neighbour[i];
    int tmp = 0, j = 0, s = v.size(), k, tree = 1, subtree;
    for(; j < s; ++j){
        k = v[j];
        if(k != pre){
            subtree = postTraverse(k, i);
            tree += subtree;
            tmp = max(tmp, subtree);
        }
    }
    tmp = max(tmp, n - tree);
    if(tmp < balance || tmp == balance && i < index){
        balance = tmp;
        index = i;
    }
    return tree;
}

int main()
{
    int test, i, x, y;
    for(scanf("%d", &test); test--; ){
        scanf("%d", &n);
    //initialize
        for(i = 1; i <= n; ++i) neighbour[i].clear();
        balance = 20002;
    //input edges
        for(i = 1; i < n; ++i){
            scanf("%d%d", &x, &y);
            neighbour[x].push_back(y);
            neighbour[y].push_back(x);
        }
    //post traverse and find each subtree's nodes count, then solve the problem
        postTraverse(1, 0);
        printf("%d %d\n", index, balance);
    }
    return 0;
}

明明省去了找leave的过程，没想到运行时间却变长了，真是……