uva 10801 - Lift Hopping

乘坐电梯 寻找最短时间的路线 即最短路问题

d[e.to]>d[u]+e.dist+60 最后输出的时候减去60 因为第一次换电梯不需要时间

// 需要注意的是k==0时 要进行特判 不然会输出-60、

#include<cstdio>

#include<cstring>

#include<vector>

#include<queue>

#include<cstdlib>

#define maxn 100001

using namespace std;

struct Edge

{

int from,to,dist;

};

struct heapnode

{

int d,u;

bool operator < (const heapnode &rhs) const

{

return d>rhs.d;

}

};

vector<Edge> edges;

vector<int> G[maxn];

bool done[maxn];

int p[maxn];

int d[maxn];

int n,num[maxn],t[10],k;

void init(int n)

{

//this->n=n;

for(int i = 0; i <= maxn; i++)

G[i].clear();

edges.clear();

}

void Addedge(int from,int to,int dist)

{

edges.push_back((Edge)

{

from,to,dist

});

int m = edges.size();

G[from].push_back(m-1);

}

void dijkstra(int s)

{

int INF = 1<<30;

priority_queue<heapnode> Q;

for(int i = 0; i <= maxn; i++)

d[i] = INF;

d[s] = 0;

memset(done,0,sizeof(done));

Q.push((heapnode)

{

0,s

});

while(!Q.empty())

{

heapnode x = Q.top();

Q.pop();

int u = x.u;

if(done[u]) continue;

done[u]=true;

for(int i = 0; i < G[u].size(); i++)

{

Edge& e = edges[G[u][i]];

if(d[e.to]>d[u]+e.dist+60)

{

d[e.to] = d[u]+e.dist+60;

p[e.to] = G[u][i];

Q.push((heapnode)

{

d[e.to],e.to

});

}

}

}

}

int build(int l,int te)

{

for(int i = 0; i < l; i++)

for(int j = 0; j < l; j++)

if(i!=j)

{

int x = abs(num[j]-num[i])*te;

Addedge(num[i],num[j],x);

}

}

int main()

{

while(scanf("%d %d",&n,&k)==2)

{

init(maxn);

for(int i = 0; i < n; i++)

scanf("%d",&t[i]);

for(int i = 0; i < n; i++)

{

int l=0;

memset(num,0,sizeof(num));

while(1)

{

scanf("%d",&num[l++]);

if(getchar()=='\n') break;

}

build(l,t[i]);

}

dijkstra(0);

if(d[k]==1<<30)

puts("IMPOSSIBLE");

else if(k==0)

printf("0\n");

else

printf("%d\n",d[k]-60);

}

return 0;

}