HDU 4522 最短路

题目描述：给出列车的行驶路径，和每个路径是硬座还是软卧，并且给出硬座和软卧的不舒适度，求从起点到终点最小的不舒适度。

思路：首先分别求出硬座和软卧从起点到终点的距离，最后比较不舒适度，当然这题得建2个图，一个是硬座的路径，一个是软卧的路径，得注意的是，当K= 1时，是硬座软卧都有，所以两边都得加进去。

其他的没什么，就是最基础的最短路。比赛的时候没有好好看题=。=

#include <iostream>#include <cstdio>#include <algorithm>#include <string>#include <cmath>#include <cstring>#include <queue>#include <set>#include <vector>#include <stack>#include <map>#include <iomanip>#define PI acos(-1.0)#define Max 2005#define inf 1<<28#define LL(x) (x<<1)#define RR(x) (x<<1|1)#define FOR(i,s,t) for(int i=(s);i<=(t);++i)#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define mp(a,b) make_pair(a,b)using namespace std;int head[2][500];struct kdq{    int s,e,next;} edge[2][2000];int num[2]  ;bool vis[2][500];void add(int s,int e,int k ){    edge[k][num[k]].e = e;    edge[k][num[k]].next = head[k][s];    head[k][s] = num[k] ++;}void init(){    mem(head,-1);    mem(vis,0);    num[0] = num[1] = 0 ;}char a[10005];int StringToInt(string x){    int l = x.size();    int num = 0;    for (int i = l - 1 ; i >= 0 ; i --)    {        num += (x[i] - '0') * pow(10.0,(double)(l - i - 1));    }    return num ;}int dis[2][500];int n ;#define x first#define y secondint spfa(int s,int e,int k){    for (int i = 0 ;i <= n ; i ++)dis[k][i] = inf;    dis[k][s] = 0;    vis[k][s] = 1;    queue<pair<int,int> >q;    q.push(mp(s,0));    while(!q.empty())    {        int temp = q.front().x;        int step = q.front().y;        q.pop();        vis[k][temp] = 0;        if(temp == e)        return step ;        for (int i = head[k][temp] ; i != -1 ;i = edge[k][i].next)        {            int tt = edge[k][i].e;            if(dis[k][tt] > dis[k][temp] + 1)            {                dis[k][tt] = dis[k][temp] + 1;                if(!vis[k][tt])                {                    vis[k][tt] = 1;                    q.push(mp(tt,step + 1));                }            }        }    }    return -1;}int main(){    int T;    cin >> T;    while ( T -- )    {        int  m ;        cin >> n >> m;        init();        while ( m -- )        {            scanf("%s",a);            int k ;            cin >> k;            int l = strlen(a);            string x ;            x.clear();            vector<int>q;            for (int i = 0 ; i < l ; i ++)            {                if(a[i] == '+')                {                    q.push_back(StringToInt(x));                    x.clear();                }                else                    x += a[i];            }            q.push_back(StringToInt(x));            l = q.size();            for (int j = 0 ; j <= k ; j ++)//当k = 1 时， 硬座和软卧都要加入该路径。                for (int i = 1 ; i < l ; i ++)                {                    add(q[i-1],q[i],j);                }            //for(int i = 0 ;i < l ;i ++)cout <<q[ i ]<<endl;            q.clear();        }        int s,e,S,D;        cin >> S>>D>>s>>e;        int step1 = spfa(s,e,0);//硬座的距离        int step2 = spfa(s,e,1);//软卧的距离        //cout <<step1 * S<<" "<<step2 * D<<endl;        if(step1 == -1 && step2 == -1)//无法到达        cout << -1<<endl;        else        {            if(step1 == -1)            cout <<step2 * D<<endl;            else if(step2 == -1)            cout <<step1 * S<<endl;            else            cout << min(step1 * S,step2 * D)<<endl;        }    }    return 0;}