toj 2870 理解dijstra

题意：求离一个城市第k远的城市的编号，无负权边。

思路：根据dijstra算法的特点可以知道，外层循环n次，每次求出的点就是离源点第i远的点的最短路。所以迭代k次，第k次新求出的点即为答案。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int INF = 9999999;
const int N = 201;
const int M = 20000;
int head[N];
int dist[N];
bool visit[N];
int n, m, k, e;

struct Edge
{
    int v, next, w;
} edge[M];

void addEdge( int u, int v, int w )
{
    edge[e].v = v;
    edge[e].w = w;
    edge[e].next = head[u];
    head[u] = e++;
}

int dij( int s )
{
    memset( visit, false, sizeof(visit) );
    for ( int i = 0; i <= n; i++ )
    {
        dist[i] = INF;
    }
    dist[s] = 0;
    int u;
    for ( int i = 0; i <= k; i++ )
    {
        u = n;
        for ( int j = 0; j < n; j++ )
        {
            if ( !visit[j] && dist[j] < dist[u] )
            {
                u = j;
            }
        }
        visit[u] = true;
        for ( int j = head[u]; j != -1; j = edge[j].next )
        {
            int v = edge[j].v, w = edge[j].w;
            if ( !visit[v] && dist[u] + w < dist[v] )
            {
                dist[v] = dist[u] + w;
            }
        }
    }
    return u;
}

int main ()
{
    while ( scanf("%d", &n), n )
    {
        scanf("%d", &m);
        e = 0;
        memset( head, -1, sizeof(head) );
        while ( m-- )
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            addEdge( u, v, w );
            addEdge( v, u, w );
        }
        scanf("%d", &k);
        printf("%d
", dij(0));
    }
    return 0;
}