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  • ACM-ICPC 2017 沈阳赛区现场赛 G. Infinite Fraction Path && HDU 6223(BFS)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6223

    参考题解:https://blog.csdn.net/qq_40482495/article/details/78492841

    注意优先队列自定义比较级的用法!!

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define ll long long
     4 #define ull unsigned long long
     5 #define mst(a,b) memset((a),(b),sizeof(a))
     6 #define mp(a,b) make_pair(a,b)
     7 #define pi acos(-1)
     8 #define pii pair<int,int>
     9 #define pb push_back
    10 const int INF = 0x3f3f3f3f;
    11 const double eps = 1e-6;
    12 const int MAXN = 15e4 + 10;
    13 const int MAXM = 2e6 + 10;
    14 
    15 char s[MAXN];
    16 int a[MAXN], ans[MAXN], vis[MAXN], nex[MAXN];
    17 
    18 struct node {
    19     int pos,val,dep;
    20 };
    21 
    22 struct cmp {
    23     bool operator()(const node &x,const node &y) {
    24         if(x.dep != y.dep) return x.dep > y.dep;
    25         else if(x.val != y.val) return x.val < y.val;
    26         return x.pos < y.pos;
    27     }
    28 };
    29 
    30 int main() {
    31 #ifdef local
    32     freopen("data.txt", "r", stdin);
    33 #endif
    34     int cas = 1;
    35     int t;
    36     scanf("%d",&t);
    37     while(t--) {
    38         int n;
    39         scanf("%d",&n);
    40         scanf("%s",s);
    41         int mx = 0;
    42         for(int i = 0; i < n; i++) {
    43             ans[i] = vis[i] = -1;
    44             a[i] = s[i] - '0';
    45             mx = max(mx, a[i]);
    46             nex[i] = (1ll * i * i + 1) % n;
    47         }
    48         priority_queue<node, vector<node>, cmp >q;
    49         for(int i = 0; i < n; i++)
    50             if(a[i] == mx) q.push({i,a[i],0});
    51         while(!q.empty()) {
    52             node now = q.top();
    53             q.pop();
    54             int pos = now.pos, val = now.val, dep = now.dep;
    55             if(dep >= n) break;
    56             if(ans[dep] > val) continue;
    57             else if(ans[dep] == val && vis[pos] == dep) continue;
    58             ans[dep] = val, vis[pos] = dep;
    59             q.push({nex[pos],a[nex[pos]],dep + 1});
    60         }
    61         printf("Case #%d: ",cas++);
    62         for(int i = 0; i < n; i++) printf("%d",ans[i]);
    63         printf("
    ");
    64     }
    65     return 0;
    66 }
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  • 原文地址:https://www.cnblogs.com/scaulok/p/9744004.html
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