hdu6805 Deliver the Cake（拆点，dijsktra）

Deliver the Cake

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1766    Accepted Submission(s): 442

Problem Description

It is Zhang3's birthday! Zhang3 has bought a birthday cake and now it's time to take it home.

There are n villages, labeled 1,2,…,n. There are m bidirectional roads, the ith of which connects village ai, bi and it is di meter(s) long.

The bakery locates at village s and Zhang3's home locates at village t. So Zhang3 wants to carry the cake from s to t. She can carry the cake either with her left hand or with her right hand. She can switch to the other hand during the trip, which takes extra x second(s) each time (when she's performing this action, she must stay in her place). Switching is allowed at any place, including the middle of the roads. She can do this as many times as she like, or don't do it at all.

Some villages are LEFT. When Zhang3 is at a LEFT village, she must carry the cake with her left hand at the moment. In the same way, some other villages are RIGHT, she must carry with her right hand when she's at these villages. The rest villages are called MIDDLE. There's no special rules at MIDDLE villages.

Zhang3 can start and finish with any hand carrying the cake. However, if s or t is not MIDDLE, their special rules must be followed.

Please help Zhang3 find a way to take the cake home, with the minimum amount of spent time.

 

Input

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

For each test case, the first line contains five integers 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 length n, describing the type of each village. The ith character is either L representing village i is LEFT, or M representing MIDDLE, or R representing RIGHT.

Finally, m lines follow, the ith of which contains three integers ai,bi,di(1≤di≤109), denoting a road connecting village ai and bi of length di.

It is guaranteed that t can be reached from s.

The sum of n in all test cases doesn't exceed 2×105. The sum of m doesn't exceed 4×105.

 

Output

For 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

 

Source

2020 Multi-University Training Contest 4

思路：

 将必须左手的村庄拆成两个左手,右手拆成两个右手，M的村庄拆成一左一右，然后跑dijsktra，然后因为是双向边，拆点后点数量变为原来2倍，连边情况变为原来的4倍，所以存边的数组要开8倍；

#include<bits/stdc++.h>
using namespace std;

#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int TT;cin>>TT;for(int tt=1;tt<=TT;tt++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repp(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define perr(i,a,b) for(int i=a;i>b;i--)
#define pb push_back
#define eb push_back
#define mst(a,b) memset(a,b,sizeof(a))
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
const int INF=0x3f3f3f3f;
const ll LINF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-12;
const double PI=acos(-1.0);
const double angcst=PI/180.0;
const ll mod=998244353;
ll max_3(ll a,ll b,ll c){if(a>b&&a>c)return a;if(b>c)return b;return c;}
ll min_3(ll a,ll b,ll c){if(a<b&&a<c)return a;if(b<c)return b;return c;}
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll qpow(ll a,ll n){ll r=1;while(n){if(n&1)r=(r*a)%mod;n>>=1;a=(a*a)%mod;}return r;}
ll qmul(ll a,ll b){ll s=(long double)a/mod*b;s=a*b-s*mod;if(s<0)s+=mod;if(s>=mod)s-=mod;return s;}


const int maxn=200010;

char ss[maxn],sss[maxn];
int n,m,s,t,x;
ll dis[maxn];
int head[maxn],tot;
struct node
{
    int u,v,nxt;
    ll w;
}e[8*maxn];
struct qnode
{
    int v;
    ll dis;
    bool operator < (const qnode &r)const
    {
        return dis>r.dis;
    }
};
void add(int u,int v,ll w)
{
    e[tot].u=u;
    e[tot].v=v;
    e[tot].w=w;
    e[tot].nxt=head[u];
    head[u]=tot++;

    e[tot].u=v;
    e[tot].v=u;
    e[tot].w=w;
    e[tot].nxt=head[v];
    head[v]=tot++;
}
void init()
{
    rep(i,1,2*n)head[i]=-1;
    tot=0;
}
ll dij(int s)
{
    rep(i,1,2*n)dis[i]=LINF;
    priority_queue<qnode>q;
    dis[s]=0;
    q.push({s,0});
    qnode tmp;
    while(!q.empty())
    {
        tmp=q.top();
        q.pop();
        int u=tmp.v;
        if(tmp.dis>dis[u])continue;
        for(int i=head[u];i!=-1;i=e[i].nxt)
        {
            int v=e[i].v;
            ll w=e[i].w;
            ll z=(sss[u]==sss[v]?0:x);
            if(dis[u]+w+z<dis[v])
            {
                dis[v]=dis[u]+w+z;
                q.push({v,dis[v]});
            }
        }
    }
    return min(dis[t],dis[t+n]);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d%d",&n,&m,&s,&t,&x);
        init();
        scanf("%s",ss);
        rep(i,1,n)
        {
            if(ss[i-1]=='M')sss[i]='L',sss[i+n]='R';
            else if(ss[i-1]=='L')sss[i]=sss[i+n]='L';
            else sss[i]=sss[i+n]='R';
        }
        while(m--)
        {
            int u,v;
            ll w;
            scanf("%d%d%lld",&u,&v,&w);
            add(u,v,w);add(u+n,v,w);
            add(u+n,v+n,w);add(u,v+n,w);
        }

        printf("%lld\n",min(dij(s),dij(s+n)));
    }
    return 0;
}