nowcoder 提高第六场A题

Solution

60分

因为所有的字母要么全相同要么全不同, 所以两条路径比较字典序只需要比较第一条边就可以, 于是建反图, 在反图上按拓扑序转移就可以.
因为有环, 所以拓扑完入度还是不为0的点答案为Infinity.

#include <queue>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const int N = 1e6 + 7;
const long long mod = 998244353;

struct Edge {
    int v, c; Edge* nxt;
    Edge(int _, int __, Edge* ___) : v(_), c(__), nxt(___) { }
};
Edge* head[N];
int du[N];

#define AddEdge(u, v, c) head[u] = new Edge(v, c, head[u])

int vis[N], dis[N], res[N], path[N], len[N];

int main () {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i += 1) {
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        AddEdge(v, u, c), du[u] += 1;
    }
    std:: queue<int> Que;
    for (int i = 1; i <= n; i += 1) 
        if (not du[i]) Que.push(i), vis[i] = true;
    while (not Que.empty()) {
        int u = Que.front(); Que.pop();
        for (auto edge = head[u]; edge; edge = edge->nxt) {
            int v = edge->v, c = edge->c;
            du[v] -= 1;
            if (not du[v]) Que.push(v), vis[v] = 1;
            if (dis[u] + 1 > dis[v]) 
                dis[v] = dis[u] + 1, path[v] = u, len[v] = c, 
                res[v] = (res[u] * 29ll % mod + c * 29ll % mod) % mod;
            if (dis[u] + 1 == dis[v] and c < len[v]) 
                path[v] = u, len[v] = c, 
                res[v] = (res[u] * 29ll % mod + c * 29ll % mod) % mod;
        }
    }
    for (int i = 1; i <= n; i += 1)
        vis[i] ? printf("%d\n", res[i]) : printf("Infinity\n");
    return 0;
}

100分-倍增

剩下的40分难点在于比较路径字典序的大小.
如何找出两条路径第一个不同的位置?
可以用倍增, 具体就是预处理出点u向后走\(1\)个位置路径的hash值, 向后走\(2\)个位置路径的hash值, 向后走\(2^k\)个位置路径的hash值.
考试的时候真的是没有想到呀.

100-分层图排序