Sicily 4832. Paper Route_n+1个节点n条边-优快云博客

本文链接：https://blog.youkuaiyun.com/Ederick/article/details/7295555

本文介绍了2012年某次编程比赛中关于寻找一棵树形结构中从任意节点到学校路径长度的最小值问题。通过深度优先搜索(DFS)算法计算各节点到根节点的距离，并利用这些距离来找到最优解。

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2012年每周一赛第一场第一题，注意到这是一棵树（因为N+1个节点只有N条边即可完全连通）。若选择第i个点作为终点，易知总长度为∑c_i-d_i+s_i（就是赛时没观察到这点），其中c是一条路径的长度，d为起点到终点的长度，s为终点到学校的长度。换言之，只需要找出s_i-d_i的最小值即可。

Run Time: 0.32sec

Run Memory: 6816KB

Code Length: 1180Bytes

Submit Time: 2012-02-26 15:06:44

// Problem#: 4832
// Submission#: 1220695
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <cstdio>
#include <vector>
using namespace std;

struct Road {
    int t;
    int dist;
    Road( int n, int d ) { t = n; dist = d; }
};

int N;
bool visited[ 100001 ] = { 0 };
int dist[ 100001 ], path[ 100001 ];
vector<Road> road[ 100001 ];

inline int min( int a, int b ) { return a < b ? a: b; }

void dfs( int n, int d ) {
    for ( vector<Road>::iterator it = road[ n ].begin(); it != road[ n ].end(); it++ ) {
        if ( !visited[ it->t ] ) {
            path[ it->t ] = d + it->dist;
            visited[ it->t ] = true;
            dfs( it->t, path[ it->t ] );
        }
    }
}

int main()
{
    int sum = 0;
    int a, b, c, i;

    scanf( "%d", &N );
    for ( i = 0; i <= N; i++ )
        scanf( "%d", &dist[ i ] );
    for ( i = 1; i <= N; i++ ) {
        scanf( "%d%d%d", &a, &b, &c );
        road[ a ].push_back( Road( b, c ) );
        road[ b ].push_back( Road( a, c ) );
        sum += c;
    }

    path[ 0 ] = 0;
    visited[ 0 ] = true;
    dfs( 0, 0 );
    int result = sum * 2 - path[ 0 ] + dist[ 0 ];
    for ( i = 1; i <= N; i++ )
        result = min( result, sum * 2 - path[ i ] + dist[ i ] );
    printf( "%d\n", result );

    return 0;

}