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.