[最短路径] HDU 1869 - 六度分离

跑N遍Dijkstra即可，时间复杂度大概是O(N3)的样子。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <assert.h>
#include <time.h>
typedef long long LL;
const int INF = 500000001;
const double EPS = 1e-9;
const double PI = acos(-1.0);
using namespace std;
int N, M;
int graph[101][101], vis[101], dis[101][101];
void init()
{
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if(i == j) graph[i][j] = 0;
else graph[i][j] = INF;
}
}
}
void Dijkstra(int s)
{
for(int i = 0; i < N; i++)
{
dis[s][i] = graph[s][i];
}
for(int i = 1; i < N; i++)
{
int k = -1;
int minn = INF;
for(int j = 0; j < N; j++)
{
if(vis[j] == -1 && minn > dis[s][j])
{
minn = dis[s][j];
k = j;
}
}
if(k == -1)
{
break;
}
vis[k] = 1;
for(int j = 0; j < N; j++)
{
if(vis[j] == -1 && dis[s][j] > dis[s][k] + graph[k][j])
{
dis[s][j] = dis[s][k] + graph[k][j];
}
}
}
}
int main()
{
#ifdef _1Test
freopen("test0.in", "r", stdin);
freopen("test0.out", "w", stdout);
srand(time(NULL));
#endif
while(~scanf("%d %d", &N, &M))
{
init();
int u, v;
for(int i = 0; i < M; i++)
{
scanf("%d %d", &u, &v);
graph[u][v] = graph[v][u] = 1;
}
for(int i = 0; i < N; i++)
{
memset(vis, -1, sizeof(vis));
vis[i] = 1;
Dijkstra(i);
}
int flag = 1;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if(dis[i][j] > 7)
{
flag = 0;
break;
}
}
if(!flag)
{
break;
}
}
if(flag) puts("Yes");
else puts("No");
}
return 0;
}