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

#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 - 1
D[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;
}