CF-938-D-最短路

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-232.CF-87-D-并查集2024-01-133.CF-91-B-单调栈+二分2024-01-234.CF-240-F-线段树2024-01-215.CF-242-E-线段树2024-01-206.CF-292-D-并查集2024-01-237.CF-282-E-Trie2024-01-178.CF-342-E-分块2024-01-129.CF-375-D-启发式合并2024-01-1210.CF-431-D-二分+数位DP2024-01-2411.CF-461-B-树形DP2024-01-2212.CF-514-D-单调队列2024-01-1413.CF-522-D-线段树2024-05-0514.CF-570-D-启发式合并2024-01-2115.CF-600-E-启发式合并2024-05-0416.CF-613-D-虚树2024-01-1317.CF-620-E-DFS序+线段树2024-01-1718.CF-675-E-RMQ优化DP2024-01-1619.CF-702-E-倍增2024-01-1720.CF-797-E-根号分治2024-05-0421.CF-817-E-Trie2024-01-1622.CF-877-E-dfs序+线段树2024-05-0523.CF-915-F-并查集2024-01-21

24.CF-938-D-最短路2024-05-10

25.CF-1051-F-最短路+最小生成树2024-01-1726.CF-1184-E3-最小生成树+倍增+并查集2024-01-2427.CF-1304-E-倍增LCA+思维2024-01-2528.CF-1399-E2-优先队列2024-01-2229.CF-1515-F-思维2024-02-2430.CF-1831-E-卡特兰数+异或哈希+差分2024-01-2231.CF-1921-F-根号分治2024-01-24

收起

938-D 题目大意

给定一张$n$个顶点$m$条边的无向图，边带权，且每个点$i$有点权$a[i]$，记$dist(i,j)$为点$i$到点$j$所有的路径中经过的最小的边权和，请求出对于每个点$i$的：

$$\underset{j}{\overset{n}{min}}(dist(i,j)+a[j])$$

---

Solution

题目涉及最短路，启发我们使用$dijkstra$求解，但对每个点求一遍最短路复杂度过高。

观察式子，除去求最短路的$dist$外，还需要附带一个点权$a[j]$。一个经典的做法是化点权为边权，我们建立一个虚拟源点，把所有节点$i$向这个虚拟源点连一条边，边权设为$a[i]$，然后从这个虚拟源点跑一遍$dijkstra$即可求出所有答案。

时间复杂度：$O(mlogm)$

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n,m;
	cin>>n>>m;
	vector<vector<pair<int,ll>>> e(n+1);
	for(int i=0;i<m;i++){
		int x,y;
		ll w;
		cin>>x>>y>>w;
		e[x].push_back({y,2LL*w});
		e[y].push_back({x,2LL*w});
	}
	for(int i=1;i<=n;i++){
		ll w;
		cin>>w;
		e[0].push_back({i,w});
	}
	vector<ll> dist(n+1,1e18);
	vector<int> vis(n+1);
	priority_queue<pair<ll,int>,vector<pair<ll,int>>,greater<>> q;
	q.push({0,0});
	dist[0]=0;
	while(!q.empty()){
		auto [d,x]=q.top();
		q.pop();
		if(vis[x]) continue;
		vis[x]=1;
		for(auto [y,w]:e[x]){
			if(d+w<dist[y]){
				dist[y]=d+w;
				q.push({dist[y],y});
			}
		}
	}
	for(int i=1;i<=n;i++){
		cout<<dist[i]<<" ";
	}
	return 0;
}