2017ACM/ICPC亚洲区沈阳站_Infinite Fraction Path(BFS)

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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define INF 0x3f3f3f3f#define rep0(i, n) for (int i = 0; i < n; i++)#define rep1(i, n) for (int i = 1; i <= n; i++)#define rep_0(i, n) for (int i = n - 1; i >= 0; i--)#define rep_1(i, n) for (int i = n; i > 0; i--)#define MAX(x, y) (((x) > (y)) ? (x) : (y))#define MIN(x, y) (((x) < (y)) ? (x) : (y))#define mem(x, y) memset(x, y, sizeof(x))#define MAXN 150000 + 10#include <queue>/**题目大意一个长度为n的数列，下标 0 ~ n - 1D[i] -> D[(i * i + 1) % n] 组成每个节点有一个出度的有向图遍历出一条长度为n的词典序最大的路径思路先使序列中最大的节点入队 用两个变量ma1 ma2 ma1 记录此层次节点的最大值(真)ma2 记录下一层节点最大值(假)当遍历完一层时 ma1 = m2用book标记同层内节点不重复入队*/using namespace std;typedef long long LL;char d[MAXN];LL n, nxt[MAXN], cnt;int ma1, ma2;queue< pair <LL, LL> > q;   ///层次 下标int book[MAXN];void work(){    LL step = -1;    while (cnt < n)    {        pair<LL, LL> tmp = q.front();        q.pop();        if (tmp.first > step)        {            step++;            ma1 = ma2;            if (cnt < n)            {                printf("%d", ma1);                cnt++;            }            ma2 = -1;            if (d[tmp.second] - '0' == ma1)            {                pair<LL, LL> t;                t.second = nxt[tmp.second];                t.first = tmp.first + 1;                q.push(t);                book[t.second] = t.first;                ma2 = d[ nxt[tmp.second] ] - '0';            }        }        else if (d[tmp.second] - '0' == ma1 && d[ nxt[tmp.second] ] - '0' >= ma2 && book[nxt[tmp.second]] != tmp.first + 1)        {            if (d[ nxt[tmp.second] ] - '0' > ma2)                ma2 = d[ nxt[tmp.second] ] - '0';            pair<LL, LL> t;            t.second = nxt[tmp.second];            t.first = tmp.first + 1;            book[t.second] = tmp.first + 1;            q.push(t);        }    }}int main(){    #ifndef ONLINE_JUDGE        freopen("in.txt", "r", stdin);    #endif // ONLINE_JUDGE    int t, kase = 0;    scanf("%d", &t);    while (t--)    {        //mem(book, 0);        scanf("%lld", &n);        scanf("%s", d);        while (!q.empty())            q.pop();        ma1 = -1;        ma2 = -1;        for (LL i = 0; i < n; i++)            book[i] = 0;        for (LL i = 0; i < n; i++)        {            nxt[i] = (i * i + 1) % n;            if (d[i] - '0' > ma2)            {                while (!q.empty())                    q.pop();                pair<LL, LL> tmp;                tmp.first = 0;                tmp.second = i;                q.push(tmp);                ma2 = d[i] - '0';            }            else if (d[i] - '0' == ma2)            {                pair<LL, LL> tmp;                tmp.first = 0;                tmp.second = i;                q.push(tmp);            }        }        cnt = 0;        printf("Case #%d: ", ++kase);        work();        printf("\n");    }    return 0;}
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