题意：

给定n个点m条边的无向图

每次必须沿着LOVE走，到终点时必须是完整的LOVE，且至少走出一个LOVE，

问这样情况下最短路是多少，在一样短情况下最多的LOVE个数是多少。

有自环。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 ll;
const ll Inf = 4611686018427387904LL;
const int N = 1314 + 100;
const int E = 13520 * 2 + 100;
const int M = N * 4 + 100;
struct Edge {
	ll len;
	int v, f, nex;
	Edge() {
	}
	Edge(int _v, int _f, ll _len, int _nex) {
		v = _v;
		f = _f;
		len = _len;
		nex = _nex;
	}
};
struct node{
	int to, f;
	node(int b=0,int d=0):to(b),f(d){}
};
Edge eg[E];
ll dis[N][4], tim[N][4];
bool vis[N][4];
int T, n, g[N], idx;

int re(char c) {
	if (c == 'L')
		return 0;
	else if (c == 'O')
		return 1;
	else if (c == 'V')
		return 2;
	else
		return 3;
}
void addedge(int u, int v, ll len, int f) {
	eg[idx] = Edge(v, f, len, g[u]);
	g[u] = idx++;
}
void spfa() {
	memset(vis, 0, sizeof vis);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < 4; ++j) {
			dis[i][j] = Inf;
			tim[i][j] = 0;
	}
	queue<node>q;
	q.push(node(0,3));
	dis[0][3] = 0;
	tim[0][3] = 0;
	while(!q.empty()){
		node u = q.front(); q.pop(); vis[u.to][u.f] = 0;
		for(int i = g[u.to]; ~i; i = eg[i].nex){
			int y = eg[i].v, f = eg[i].f;
			if(f != (u.f+1)%4)continue;
			bool yes = false;
			if(dis[y][f] > dis[u.to][u.f]+eg[i].len)
			{
				dis[y][f] = dis[u.to][u.f]+eg[i].len;
				tim[y][f] = tim[u.to][u.f];
				if(f == 3)
					tim[y][f]++;
				yes = true;
			}
			else if(dis[y][f] == dis[u.to][u.f]+eg[i].len) {
				ll tmp = tim[u.to][u.f];
				if(f == 3)
					tmp++;
				if(tmp > tim[y][f])
					tim[y][f] = tmp, yes = true;
			}
			else if(tim[y][f]==0) {
				ll tmp = tim[u.to][u.f];
				if(f == 3)
					tmp++;
				if(tmp > tim[y][f])
					dis[y][f] = dis[u.to][u.f]+eg[i].len, tim[y][f] = tmp, yes = true;
			}
			if(yes && vis[y][f] == 0)
				vis[y][f] = 1, q.push(node(y, f));
		}
	}
}
void work() {
	int m, u, v; ll len;
	char s[5];
	memset(g, -1, sizeof g);
	idx = 0;

	scanf("%d %d", &n, &m);
	while (m -- > 0) {
		scanf("%d%d%I64d%s", &u, &v, &len, s);
		-- u; -- v;
		addedge(u, v, len, re(s[0]));
		addedge(v, u, len, re(s[0]));
	}
	spfa();
	ll ansdis = dis[n - 1][3], ansnum = tim[n - 1][3];
	printf("Case %d: ", ++T);
	if (ansdis == Inf || ansnum == 0) {
		puts("Binbin you disappoint Sangsang again, damn it!");
	} else {
printf("Cute Sangsang, Binbin will come with a donkey after travelling %I64d meters and finding %I64d LOVE strings at last.\n", ansdis, ansnum);
	}
}
int main() {
	int cas;
	T = 0;
	scanf("%d", &cas);
	while (cas -- > 0)
		work();
	return 0;
}
/*
99
4 4
1 2 1 L
2 4 1 O
4 1 1 V
1 4 1 E

1 4
1 1 1 L
1 1 1 O
1 1 1 V
1 1 1 E

1 0


*/