[codevs1391]伊吹萃香

题目←

东方系列的可做题……
每点有原色 -> 反色两种状态，所以把数组开双倍大
i表示原色，i+n表示与原色颜色相反
注意在路上走也是消耗时间的，到达时目标点已经变色了
建好图之后暴力跑最短路就可以了，最后在n和n + n里取min
其实spfa和dij都能过，一开始T掉是因为数组开小了……

#include
#include
#include
#include
#include
#define INF 2147483645
using namespace std;
const int MAXN = 100000 + 50;
int head[MAXN],next[MAXN << 1],tot,dis[MAXN];
struct edge
{
    int f,t,v;
}l[MAXN << 1];
void init(int n)
{
    for(int i = 1;i <= n;i ++)
    {
        head[i] = -1;
        dis[i] = INF;
    }
}
void build(int f,int t,int v)
{
    l[++ tot] = (edge){f,t,v};
    next[tot] = head[f];
    head[f] = tot;
}
int n,m;
bool co[MAXN];
int w[MAXN],s[MAXN],a,b,c;
struct zt
{
    int num,dis;
};
bool operator < (zt a,zt b)
{
    return a.dis > b.dis;
}
priority_queue  q;
//queue <int> q;
bool used[MAXN];
void dij(int x)
{
    dis[x] = 0;
    q.push((zt){x,dis[x]});
    while(!q.empty())
    {
        zt u = q.top();
        q.pop();
        used[u.num] = true;
        if(used[n] && used[n + n])return;
        for(int i = head[u.num];i != -1;i = next[i])
        {
            int t = l[i].t;
            if(dis[t] > dis[u.num] + l[i].v)
            {
                dis[t] = dis[u.num] + l[i].v;
                q.push((zt){t,dis[t]});
            }
        }
    }
}
/*void spfa(int x)
{
    dis[x] = 0;
    used[x] = true;
    q.push(x);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        used[u] = false;
        for(int i = head[u];i != -1;i = next[i])
        {
            int t = l[i].t;
            if(dis[t] > dis[u] + l[i].v)
            {
                dis[t] = dis[u] + l[i].v;
                if(!used[t])q.push(t);
                used[t] = true;
            }
        }
    }
}*/
int main()
{
    scanf("%d%d",&n,&m);
    init(n << 1 | 1);
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&co[i]);
    }
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&w[i]);
    }
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&s[i]);
    }
    for(int i = 1;i <= m;i ++)
    {
        scanf("%d%d%d",&a,&b,&c);
        int del = abs(w[a] - w[b]);
        if((co[a] && co[b]) || (!co[a] && !co[b]))//手滑打成!co[a]&&co[b],wa了两次
        {
            build(a,b + n,c);
            build(a + n,b,c);
        }
        else if(co[a] && !co[b])
        {
            build(a,b + n,c + del);
            build(a + n,b,max(0,c - del)); 
        }
        else if(!co[a] && co[b])
        {
            build(a,b + n,max(0,c - del));
            build(a + n,b,c + del);
        }
    }
    for(int i = 1;i <= n;i ++)
    {
        if(!co[i])
        {
            build(i,i + n,0);
            build(i + n,i,s[i]);
        }
        else
        {
            build(i,i + n,s[i]);
            build(i + n,i,0);
        }
    }
    dij(1);
    //spfa(1);
    printf("%d",min(dis[n],dis[n + n]));
    return 0;
}