uva 10801【Lift Hopping】

表示对这类题目没feel，抽象。。。参考别人的代码后写出来的，顺便练习了一下FIFO队列实现Bellman-ford算法。。。

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;

int g[6][105];
int n,k;
int t[6];
int d[105];
int v[105];

void init()
{
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&t[i]);
    }

    memset(g,0,sizeof(g));
    char ch;
    int a;
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&a);
        g[i][a] = 1;
        int b;
        while((ch = getchar()) != '\n')
        {
            scanf("%d",&b);
            g[i][b] = 1;
        }
    }
}

void solve()
{
    memset(v,0,sizeof(v));
    memset(d,-1,sizeof(d));
    d[0] = 0;
    queue<int> q;
    q.push(0);
    v[0] = 1;
    while(!q.empty())
    {
        int x = q.front();
        q.pop();
        v[x] = 0;
        for(int i = 1;i <= n;i ++)
        {
            if(g[i][x])
            {
                for(int j = 1;j <= 100;j ++)
                {
                    if(g[i][j])
                    {
                        int temp = d[x] + t[i] * abs(j - x) + 60;
                         if(d[j] == -1 || temp < d[j])
                        {
                            d[j] = temp;
                            if(!v[j])
                            {
                                v[j] = true;
                                q.push(j);
                            }
                        }
                    }
                }
            }
        }
    }

    if(d[k] == -1)
    {
        printf("IMPOSSIBLE\n");
    }
    else if(k == 0)
    {
        printf("0\n");
    }
    else
    {
        printf("%d\n",d[k]-60);
    }
}

int main()
{
    while(scanf("%d%d",&n,&k) == 2)
    {
        init();
        solve();
    }

    return 0;
}