描述
题解
floyd搞搞事情,求任意两点之间的距离是否超过7,默认每条路径长度为1。
代码
#include <iostream>
#include <cstring>
using namespace std;
/*
* Floyd算法,求从任意节点i到任意节点j的最短路径
* cost[][]:初始化为INF(cost[i][i]:初始化为0)
*/
const int MAXN = 100;
int cost[MAXN][MAXN];
int lowcost[MAXN][MAXN];
int path[MAXN][MAXN];
void Floyd(int n)
{
memcpy(lowcost, cost, sizeof(cost));
memset(path, -1, sizeof(path));
for (int k = 0; k < n; k++)
{
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (lowcost[i][j] > (lowcost[i][k] + lowcost[k][j]))
{
lowcost[i][j] = lowcost[i][k] + lowcost[k][j];
path[i][j] = k;
}
}
}
}
return ;
}
int main(int argc, const char * argv[])
{
int N, M;
while (cin >> N >> M)
{
memset(cost, 0x3f, sizeof(cost));
int a, b;
for (int i = 0; i < M; i++)
{
scanf("%d%d", &a, &b);
cost[a][b] = cost[b][a] = 1;
}
Floyd(N);
bool flag = true;
for (int i = 0; i < N; i++)
{
for (int j = i + 1; j < N; j++)
{
if (lowcost[i][j] > 7)
{
flag = false;
break;
}
}
}
if (flag)
{
cout << "Yes\n";
}
else
{
cout << "No\n";
}
}
return 0;
}