POJ 2949 Word Rings
传送门
同 AcWing 1165 单词环、洛谷 SP2885 WORDRING - Word Rings、LOJ 10082 Word Rings
方法类似 Sightseeing Cows
,通过移项构造环路判定,二分搜索答案缩小目标精度,注意使用 tot >= n * 2
或 stack
代替 queue
玄学优化。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int, int> p;
const int maxn = 26 * 26;
const int maxm = 1e5 + 10;
char str[maxm];
bool vis[maxn];
double dis[maxn];
int ecnt, head[maxn], cnt[maxn];
struct edge {
int to, wt, nxt;
} edges[maxm];
template<typename T>
inline const T read()
{
T x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + ch - '0';
ch = getchar();
}
return x * f;
}
template<typename T>
inline void write(T x, bool ln)
{
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) write(x / 10, false);
putchar(x % 10 + '0');
if (ln) putchar(10);
}
inline int getId(char a, char b)
{
return (a - 'a') * 26 + b - 'a';
}
void addEdge(int u, int v, int w)
{
edges[ecnt].to = v;
edges[ecnt].wt = w;
edges[ecnt].nxt = head[u];
head[u] = ecnt++;
}
bool spfa(int n, double val)
{
queue<int> q;
memset(cnt, 0, sizeof cnt);
memset(vis, false, sizeof vis);
for (int i = 0; i < maxn; ++i) {
dis[i] = 0;
vis[i] = true;
q.push(i);
}
int tot = 0;
while (not q.empty()) {
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; compl i; i = edges[i].nxt) {
int v = edges[i].to, w = edges[i].wt;
if (dis[v] < dis[u] + w - val) {
dis[v] = dis[u] + w - val;
if (not vis[v]) {
vis[v] = true;
q.push(v);
cnt[v] = cnt[u] + 1;
if (cnt[v] >= n or ++tot > n * 2) {
return true;
}
}
}
}
}
return false;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("input.txt", "r", stdin);
#endif
while (true) {
int n = read<int>();
if (not n) break;
ecnt = 0;
memset(head, -1, sizeof head);
for (int i = 0; i < n; ++i) {
scanf("%s", str);
int len = (int)strlen(str);
if (len < 2) continue;
int u = getId(str[0], str[1]), v = getId(str[len - 2], str[len - 1]);
addEdge(u, v, len);
}
double low = 0, high = 1e3;
while (high - low > 1e-4) {
double mid = (low + high) / 2;
if (spfa(n, mid)) {
low = mid;
} else {
high = mid;
}
}
if (low == 0.0) {
puts("No solution.");
} else {
printf("%.2f\n", low);
}
}
return 0;
}