Beehives

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题意：给出一个无向图，求最短的环
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解法：由于点数只有500， 对于每一个点做一次bfs就好了，只要搜到之前更新过的点那么就是一个环了，bfs保证了最短。
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#include <iostream>#include <stdio.h>#include <stdlib.h>#include <vector>#include <algorithm>#include <string.h>using namespace std;const int maxn = 600;const int INF = 1000000;const int maxm = 21000;int dist[maxn], n, m, d[maxn], x, y, ans, pre[maxn];int tar[2*maxm], nextt[2*maxm], last[maxn], tot;template <typename T>inline bool scan_d(T &ret) {    char c; int sgn;    if (c = getchar(), c == EOF) return 0;    while (c != '-' && (c < '0' || c > '9')) c = getchar();    sgn = (c == '-')?-1:1;    ret = (c == '-')?0:(c-'0');    while (c = getchar(), c >= '0' && c <= '9') ret = ret*10+(c-'0');    ret *= sgn;    return 1;}void insert(int x, int y) {    tot++;    tar[tot] = y;    nextt[tot] = last[x];    last[x] = tot;}int bfs(int s) {    memset(dist, 255, sizeof(dist));    memset(pre, 0, sizeof(pre));    dist[s] = 0;    int i = 0, j = 1;    d[1] = s;    int minn = INF;    while (i != j) {        i++;        if (i > INF) i = 1;        int k = last[d[i]];        while (k != 0) {            if (dist[tar[k]] == -1) {                dist[tar[k]] = dist[d[i]]+1;                pre[tar[k]] = d[i];                j++;                if (j > INF) j = 1;                d[j] = tar[k];            } else if (tar[k] != pre[d[i]]) {                minn = min(minn, dist[tar[k]]+dist[d[i]]+1);            }            k = nextt[k];        }    }    return minn;}int main() {    int tt;    scanf("%d", &tt);    for (int cases = 1; cases <= tt; cases++) {        tot = 0;        memset(last, 0, sizeof(last));        scanf("%d %d", &n, &m);        for (int i = 1; i <= m; i++) {            scan_d(x);            scan_d(y);            x++; y++;            insert(x, y);            insert(y, x);        }        ans = INF;        for (int i = 1; i <= n; i++) {            ans = min(ans, bfs(i));        }        printf("Case %d: ", cases);        if (ans == INF) printf("impossible\n");        else printf("%d\n", ans);    }}