AtCoder Regular Contest 088 D - People on a Line（DFS+图论）

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原创最新推荐文章于 2026-05-18 05:51:30 发布 377 阅读 AI 写同款· GEO 优化 ›

r3live 使用前提雷达-相机外参标定 livox_camera_lidar_calibration 港科大livox_camera_calib虽然操作方便，但是使用mid360雷达会有视角问题（投影三维点到相机），尝试了很多场景，标定效果都不理想（推荐场景是楼梯间），看来港大的更适合avia之类窄角度的雷达。如果标定效果不好的话，就使用 getExt２节点，getExt1节点只优化外参，而getExt2节点在计算的时候会将一开始计算的内参作为初值和外参一起优化。一般使用getExt1节点即可，如果在外参初值验证过，并且异常值已经剔除后，优化还是有较大的残差，那么可以使用getExt2试一试。阅读详情

AtCoder Regular Contest 088 D - People on a Line（DFS+图论）

Problem Statement

There are N people standing on the x-axis. Let the coordinate of Person i be x_i. For every i, x_i is an integer between 0 and 10⁹ (inclusive). It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people. The i-th piece of information has the form (L_i,R_i,D_i). This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i}−x_{L_i}=D_i holds.

It turns out that some of these M pieces of information may be incorrect. Determine if there exists a set of values (x₁,x₂,…,x_N) that is consistent with the given pieces of information.

Constraints

1≤N≤100 000
0≤M≤200 000
1≤L_i,R_i≤N (1≤i≤M)
0≤D_i≤10 000 (1≤i≤M)
L_i≠R_i (1≤i≤M)
If i≠j, then (L_i,R_i)≠(L_j,R_j) and (L_i,R_i)≠(R_j,L_j).
D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L₁ R₁ D₁
L₂ R₂ D₂
:
L_M R_M D_M

Output

If there exists a set of values (x₁,x₂,…,x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

Sample Output 1

Yes

Some possible sets of values (x₁,x₂,x₃) are (0,1,2) and (101,102,103).

Sample Input 2

Sample Output 2

No

If the first two pieces of information are correct, x₃−x₁=2 holds, which is contradictory to the last piece of information.

Sample Input 3

Sample Output 3

Yes

Sample Input 4

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

用邻接表储存图，然后遍历，记录每个点之间距离，如果再次到达访问的点，发现距离和之前储存的不同则返回false

最后返回true；

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int N = 1e5+10;
int n,m;
int dis[N];
bool vis[N];
vector<pair<int,int> >E[N];
bool dfs(int u){
    vis[u] = 1;
    vector<pair<int,int> >::iterator it;
    for(it = E[u].begin(); it != E[u].end(); it++){
        int v = (*it).first;//注意这里的迭代器要加（*）
        int d = (*it).second;
        if(!vis[v]){
            dis[v] = dis[u] + d;
            if(!dfs(v))
                return false;
        }
        else if(dis[u]+d!=dis[v])
            return false;
    }
    return true;
}
int main(){
    cin >> n >> m;
    int u,v,w;
    int i;
    for(i = 1; i <= m; i++){
        cin >> u >> v >> w;
        E[u].push_back(make_pair(v,w));
        E[v].push_back(make_pair(u,-w));
    }
    for(i = 1; i <= n; i++){
        if(!vis[i]){
            if(!dfs(i)){
                cout << "No" << endl;
                return 0;
            }
        }
    }
    cout << "Yes" << endl;
    return 0;
}