hdu6166 Senior Pan
Senior Pan
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1411 Accepted Submission(s): 558
The task is simple : ZKC will give Pan a directed graph every time, and selects some nodes from that graph, you can calculate the minimum distance of every pair of nodes chosen in these nodes and now ZKC only cares about the minimum among them. That is still too hard for poor Pan, so he asks you for help.
Then m lines follow. Each line contains three integers xi,yi representing an edge, and vi representing its length.1≤xi,yi≤n,1≤vi≤100000
Then one line contains one integer K, the number of nodes that Master Dong selects out.1≤K≤n
The following line contains K unique integers ai, the nodes that Master Dong selects out.1≤ai≤n,ai!=aj
5 6
1 2 1
2 3 3
3 1 3
2 5 1
2 4 2
4 3 1
3
1 3 5
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; typedef long long ll;
const int maxn = ;
const ll inf = 1e17; int T,cas,n,vis[maxn],m,head[maxn],to[maxn],nextt[maxn],w[maxn],tot = ,K,a[maxn],mark[maxn];
ll d[maxn],ans = inf; struct node
{
int x;
ll len;
bool operator < (const node& a) const {
return len > a.len;
}
};
priority_queue<node> q; void add(int x,int y,int z)
{
w[tot] = z;
to[tot] = y;
nextt[tot] = head[x];
head[x] = tot++;
} ll dijkstra()
{
while (!q.empty())
{
node u = q.top();
q.pop();
int x = u.x;
ll len = u.len;
if (mark[x])
return len;
if (vis[x])
continue;
vis[x] = ;
for (int i = head[x];i;i = nextt[i])
{
int v = to[i];
if (d[v] > d[x] + w[i])
{
d[v] = d[x] + w[i];
node temp;
temp.x = v;
temp.len = d[v];
q.push(temp);
}
}
}
return inf;
} void solve()
{
for (int i = ; i < ; i++)
{
while (!q.empty())
q.pop();
memset(vis,,sizeof(vis));
memset(mark,,sizeof(mark));
memset(d,/,sizeof(d));
for (int j = ; j <= K; j++)
{
if (( << i) & a[j])
mark[a[j]] = ;
else
{
node temp;
temp.x = a[j];
temp.len = ;
d[a[j]] = ;
q.push(temp);
}
}
ans = min(ans,dijkstra());
memset(vis,,sizeof(vis));
memset(mark,,sizeof(mark));
memset(d,/,sizeof(d));
while (!q.empty())
q.pop();
for (int j = ; j <= K; j++)
{
if (( << i) & a[j])
{
node temp;
temp.len = ;
temp.x = a[j];
d[a[j]] = ;
q.push(temp);
}
else
mark[a[j]] = ;
}
ans = min(ans,dijkstra());
}
} int main()
{
scanf("%d",&T);
while (T--)
{
ans = inf;
tot = ;
memset(head,,sizeof(head));
scanf("%d%d",&n,&m);
for (int i = ; i <= m; i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
add(x,y,z);
}
scanf("%d",&K);
for(int i = ; i <= K; i++)
scanf("%d",&a[i]);
solve();
printf("Case #%d: %lld\n",++cas,ans);
} return ;
}
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