【CodeChef - CLIQUED 】Bear and Clique Distances(建图,缩点技巧,思维)
作者:互联网
题干:
解题报告:
主要就是在于怎么处理那个前K个点:组成一个团。换句话说,缩成一个点。先直接当成每个点多了k条边来处理,T了。想想也是啊,要是K=1e5,那就是1e10条边了。。刚开始尝试了半天缩点,后来发现其实不用,只需要把这k个点都连到一个新点上,最后再连回去,然后直接Dijkstra就可以了 。只是相当于多了2*k条边而已。。真是好题,,,
AC代码:
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
const ll INF = 0x3f3f3f3f3f;
int n,m,tot;
int head[MAX];
bool vis[MAX];
int k;
ll X;
ll dis[MAX];
struct Edge {
int to;
ll w;
int ne;
} e[MAX<<2];
struct Point {
int pos;
ll c;
Point() {}
Point(int pos,ll c):pos(pos),c(c) {}
bool operator < (const Point b) const {
return c > b.c;
}
};
void add(int u,int v,ll w) {
e[++tot].to = v;
e[tot].w = w;
e[tot].ne = head[u];
head[u] = tot;
}
void Dijkstra(int st) {
for(int i = 1; i<=n+1; i++) {
dis[i] = INF;
vis[i] = 0;
}
// for(int i = 1; i<=k; i++) dis[i] = X;
dis[st] = 0;
priority_queue<Point> pq;
pq.push(Point(st,dis[st]));
while(pq.size()) {
Point cur = pq.top();
pq.pop();
if(vis[cur.pos]) continue;
vis[cur.pos] = 1;
for(int i = head[cur.pos]; i!=-1; i = e[i].ne) {
if(dis[e[i].to] > dis[cur.pos] + e[i].w) {
dis[e[i].to] = dis[cur.pos] + e[i].w;
pq.push(Point(e[i].to,dis[e[i].to]));
}
}
}
}
int main()
{
int t;
cin>>t;
while(t--) {
int st;
scanf("%d%d%lld%d%d",&n,&k,&X,&m,&st);
//init
tot=0;
for(int i = 1; i<=n+1; i++) head[i] = -1;
ll c;
for(int i = 1; i<=k; i++) {
add(i,n+1,0);
add(n+1,i,X);
}
for(int a,b,i = 1; i<=m; i++) {
scanf("%d%d%lld",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
Dijkstra(st);
for(int i = 1; i<=n; i++) printf("%lld%c",dis[i],i == n ? '\n' : ' ');
}
return 0 ;
}
// if(cur.pos<=k) {
// for(int i = 1; i<=k; i++) {
// if(vis[i]) continue;
// if(dis[cur.pos] + X < dis[i]) {
// dis[i] = dis[cur.pos] + X;
// pq.push(Point(i,dis[i]));
// }
// }
// }
//// if(e[i].to <= k) {
//// for(int i = head[n+1]; i!=-1; i = e[i].ne) {
//// if(dis[e[i].to] > dis[n+1] + e[i].w+X) {
//// dis[e[i].to] = dis[n+1] + e[i].w+X;
//// pq.push(Point(e[i].to,dis[e[i].to]));
//// }
//// }
//// }
// }
// if(cur.pos<=k) {
// for(int i = head[n+1]; i!=-1; i = e[i].ne) {
// if(e[i].to <= k) {
// if(dis[e[i].to] > dis[cur.pos] + X) {
// dis[e[i].to] = dis[cur.pos] + X;
// pq.push(Point(e[i].to,dis[e[i].to]));
// }
// }
// else if(dis[e[i].to] > dis[cur.pos] + e[i].w+X) {
// dis[e[i].to] = dis[cur.pos] + e[i].w+X;
// pq.push(Point(e[i].to,dis[e[i].to]));
// }
// }
// }
// }
标签:缩点,Distances,pq,cur,int,pos,Bear,include,dis 来源: https://blog.csdn.net/qq_41289920/article/details/89061511