題意:
給定n, m, k (n <= 20, k <= n),表示有n個人,m個關係(關係爲a和b是否認識), 編號爲1-n, 1號想認識n號,可是,兩個人如果想建立關係必須滿足,他們的共同朋友大於或等於k, 求1想認識n,1最少需要認識幾個人。
題解:
給每個人與其他人的關係進行狀態壓縮,然後直接bfs搜索即可。
代碼:
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 20 + 5;
int n, m, k, rel[maxn], vis[1<<21];
struct Way{
int v, step;
Way(int v = 0, int step = 0) : v(v), step(step) {}
}str, now, nxt;
void read_input()
{
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < m; i++)
{
int x, y; scanf("%d%d", &x, &y);
x--; y--;
rel[x] |= (1 << y);
rel[y] |= (1 << x);
}
str = Way(rel[0], 0);
}
void init()
{
memset(rel, 0, sizeof(rel));
memset(vis, 0, sizeof(vis));
}
bool ok(int status)
{
int cnt = 0;
for (int i = 0; i < n; i++) {
if (status & (1<<i)) {
cnt ++;
}
}
if (cnt >= k) return true;
return false;
}
int bfs()
{
if (str.v & (1<<(n-1))) {
return 0;
}
queue<Way> Q; Q.push(str);
vis[str.v] = 1;
while (!Q.empty())
{
now = Q.front(); Q.pop();
if (ok(now.v & rel[n-1]))
{
return now.step;
}
for (int i = 0; i < n; i++)
{
if (now.v & (1 << i)) continue;
nxt = now; nxt.step++;
if (ok(now.v & rel[i]))
{
nxt.v |= (1<<i);
if (!vis[nxt.v])
{
Q.push(nxt);
vis[nxt.v] = 1;
}
}
}
}
return -1;
}
int main()
{
// freopen("/Users/apple/Desktop/in.txt", "r", stdin);
int t, kase = 0; scanf("%d", &t);
while (t--)
{
init();
read_input();
printf("Case #%d: %d\n", ++kase, bfs());
}
return 0;
}