[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<iostream>#include<cstdio>#include<algorithm>#include<queue>#include<cmath>#define INF 2147483645using 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 <zt> 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;}