https://vjudge.net/contest/386568#problem/D

InputThe first line of the input gives the number of test cases,T(1≤T≤100)T(1≤T≤100).TTtest cases follow.

For each test case, the first line contains five integersn,m,s,t,x(1≤n≤105,1≤m≤2×105,1≤x≤109)n,m,s,t,x(1≤n≤105,1≤m≤2×105,1≤x≤109), representing the number of villages, the number of roads, the bakery's location, home's location, and the time spent for each switching.

The next line contains a string of lengthnn, describing the type of each village. Theithithcharacter is eitherLLrepresenting villageiiis LEFT, orMMrepresenting MIDDLE, orRRrepresenting RIGHT.

Finally,mmlines follow, theithithof which contains three integersai,bi,di(1≤di≤109)ai,bi,di(1≤di≤109), denoting a road connecting villageaiaiandbibiof lengthdidi.

It is guaranteed thatttcan be reached fromss.

The sum ofnnin all test cases doesn't exceed2×1052×105. The sum ofmmdoesn't exceed4×1054×105.
OutputFor each test case, print a line with an integer, representing the minimum amount of spent time (in seconds).
Sample Input

1
3 3 1 3 100
LRM
1 2 10
2 3 10
1 3 100

Sample Output

100

Sponsor

參考程式碼：https://www.cnblogs.com/chuliyou/p/13416544.html

思路：

　　區別左手右手

　　分成兩個方向用dijsktra？（未解決

附草稿程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=4e5+10;
const int mod=1e9+7;

int n, m, s, t, xx;
struct Node
{
	int head;
	int ver;
	int edge;
	int Next;
}line[N*2];
bool v[N*2];
long long d[N*2];
int hand[N*2];

int tot = 0;
priority_queue <pair<long long , int> >q;
void add(int x , int y , int z)
{
	tot += 1;
    line[tot].ver = y;
    line[tot].edge = z;
    line[tot].Next = line[x].head;
    line[x].head = tot;
}
long long dijkstra(int s, int t)
{
	long long ans = 1e18;
    for(int i = 1; i<=2*n;i++)
	{
	 	d[i] = 1e18;
    }
	memset(v , 0 , sizeof(v));
    d[s] = 0;
    q.push(make_pair(0 , s));
    int i = 0 , j = 0;
    while(q.size())
    {
        int x = q.top().second;
        q.pop();
        if(v[x])
		{
		 	continue;
        }
		v[x] = 1;
        for(i = line[x].head;i != 0; i=line[i].Next)
        {
            int y = line[i].ver , z = line[i].edge;
            if(hand[x] != hand[y])
			{
			 	z += xx;//是否要換手 
            }if(d[y] > d[x]+z)
            {
                d[y] = d[x] + 1ll*z;
                q.push(make_pair(-d[y] , y));
            }
        }
    }
	return min(d[t] , d[t + n]);
}
signed main()
{
    int T = 0;
    cin >> T;
    while(T--)
    {
    	tot = 0;
    	for(int i = 1;i<=2*n;i++)
		{
		 	line[i].head = 0;
		}
    	cin >> n >> m >> s >> t >> xx;
    	string S;
    	cin >> s;
    	for(int i = 0;i<S.size();i++)
    	{
    		if(S[i] == 'L')
    		{
    			hand[i+1] = 0;
	 			hand[i+1+n] = 0;
			}
			if(S[i] == 'M')
			{
				hand[i+1] = 0;
				hand[i+1+n] = 1;
			}
			if(S[i] == 'R')
    		{
    			hand[i+1] = 1;
				hand[i+1+n] = 1;
			}
		}
		for(int i = 1; i <= m; i++)
		{
			int u = 0 , v = 0 , w = 0;
			cin >> u >> v >> w;
			add(u , v , w);
			add(v , u , w);
			add(u+n , v , w);
			add(v , u+n , w);
			add(u , v+n , w);
			add(v+n , u , w);
			add(u+n , v+n , w);
			add(v+n , u+n , w);
		}
		cout << min(dijkstra(s,t) , dijkstra(s+n, t)) <<endl;
	}
    return 0;
}