AtCoder Regular Contest 090 D People on a Line【dfs + bfs】

D - People on a Line

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

There are N people standing on the x-axis. Let the coordinate of Person i be xi. For every i, xi 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 (Li,Ri,Di). This means that Person Ri is to the right of Person Li by Di units of distance, that is, xRi−xLi=Di 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,…,xN) that is consistent with the given pieces of information.

Constraints

1≤N≤100 000
0≤M≤200 000
1≤Li,Ri≤N (1≤i≤M)
0≤Di≤10 000 (1≤i≤M)
Li≠Ri (1≤i≤M)
If i≠j, then (Li,Ri)≠(Lj,Rj) and (Li,Ri)≠(Rj,Lj).
Di are integers.

Input

Input is given from Standard Input in the following format:

N M
L1 R1 D1
L2 R2 D2
:
LM RM DM

Output

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

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

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

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

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

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes
题意： n个人，给你m个关系，n个人站在x轴上，给你的关系包括 Li,Ri,Di L i , R i , D i ，表示站在 Li L i 位置的人距离右边 Di D i 距离是 Ri R i ，让你判断这m个关系是否完全正确

分析： 首先我们考虑用vector建邻接表，对于每个点进行DFS（考虑不是连通图的情况，可能会漏判），我们可以判断进而更新新的数据，当然得利用vis标记下，BFS一样，我们可以多点一起判断，这样速度快点，具体看代码

DFS代码

#include

using namespace std;

#define ll long long
#define fi first
#define se second

const int maxn = 1e6 + 10, INF = 1000000009;

ll a[maxn];
bool flg = false;
vectorint,ll> > E[maxn];

void init() {
    for(int i = 0;i < maxn;i++) a[i] = INF;
}
int n,m;
bool vis[maxn];

void dfs(int idx) {
    for(int i = 0;i < E[idx].size();i++) {
        int to = E[idx][i].fi;
        int v = E[idx][i].se;
        if(vis[to]) continue;
        if(a[to] != INF) {
            if(a[to] - a[idx] != v) {
                flg = true;
                return ;
            }
        } else {
            a[to] = a[idx] + v;
            dfs(to);
            if(flg) return;
        }
    }
    vis[idx] = true;
    if(flg) return;

}
int L[maxn];
int main(){
    ios_base::sync_with_stdio(0);

    cin>>n>>m;
    for(int i = 0;i < m;i++) {
        int l,r,d;cin>>l>>r>>d;
        L[i] = l;
        E[l].push_back(make_pair(r,d));
        E[r].push_back(make_pair(l,-d));
    }
    init();
    for(int i = 0;i < n;i++) {
        if(E[L[i]].size() >= 1) {
            if(vis[L[i]]) continue;
            a[L[i]] = 0;
            dfs(L[i]);
            if(flg) break;
        }
    }
    if(flg) cout<<"No"<else cout<<"Yes"<return 0;
}

BFS代码

#include

using namespace std;

#define ll long long

const int maxn = 5e5 + 10, INF = 1000000009;
vectorint,int> > E[maxn];
ll d[maxn];

int main(){
    ios_base::sync_with_stdio(0);
    init();
    int n,m;cin>>n>>m;
    for(int i = 0;i < m;i++) {
        int l,r,d;cin>>l>>r>>d;
        E[r].push_back(make_pair(l,-d));
        E[l].push_back(make_pair(r,d));
    }
    for(int i = 0;i < maxn;i++) d[i] = INF;
    for(int i = 1;i <= n;i++) {
        if(d[i] == INF) {
            queue<int> q;
            q.push(i);
            d[i] = 0;
            while (!q.empty()) {
                int u = q.front();
                q.pop();
                for(int j = 0;j < E[u].size();j++) {
                    int v = E[u][j].first;
                    int val = E[u][j].second;
                    if(d[v] == INF) {
                        d[v] = d[u] + (ll)val;
                        q.push(v);
                    } else if(d[v] - d[u] != val) {
                        cout<<"No"<return 0;
                    }
                }
            }
        }
    }
    cout<<"Yes"<return 0;
}

• 如有错误或遗漏，请私聊下UP，thx