多起点最短路
题目链接
刚开始想跑n次最短路,结果 tle,其实只要跑k次最短路然后相加即可,
另外建一个点,并且和所以种类为i的建筑建一条长度为0的边,然后跑最短路即可。
代码
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 5e4+7;
const int mod = 1e9+7;
int vis[maxn];
int pace[maxn];
ll m,n,k;
int a[maxn];
struct edge
{
int to,cost;
edge(int _to,int _cost):to(_to),cost(_cost) {}
};
typedef pair<ll,ll> P;
vector<edge> G[maxn];
vector<int>aa[maxn];
ll d[maxn];
ll dis[maxn];
inline int mins(int a,int b)
{
return a<b?a:b;
};
void dijkstra(int s)
{
priority_queue<P,vector<P>,greater<P> > que;
memset(d,inf,sizeof d);
d[s]=0;
que.push(P(0,s));
while(!que.empty())
{
P p = que.top();
que.pop();
ll v = p.second;
if (d[v] < p.first)
continue;
for(int i = 0; i < G[v].size(); i++)
{
edge e = G[v][i];
if (d[e.to] > d[v] + e.cost)
{
d[e.to] = d[v] + e.cost;
que.push(P(d[e.to],e.to));
}
}
}
for(int i=1;i<=n;i++) {
dis[i]+=d[i];
}
}
int main()
{
cin>>n>>m>>k;
for(int i=1; i<=n; i++)
{
cin>>a[i];
aa[a[i]].push_back(i);
}
int x,y;
for(int i=0; i<m; i++)
{
cin>>x>>y;
G[x].push_back(edge(y,1));
G[y].push_back(edge(x,1));
}
for(int i=1;i<=k;i++){
for(int j=0;j<aa[i].size();j++){
G[0].push_back(edge(aa[i][j],0));
}
dijkstra(0);
G[0].erase(G[0].begin(),G[0].end());
}
for(int i=1;i<=n;i++) {
if(dis[i]>=inf||dis[i]<=-inf) printf("0 ");
else printf("%lld ",dis[i]);
}
return 0;
}