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HDU 5876 Sparse Graph BFS 最短路

2024-08-13 06:27:20   210人阅读

Sparse Graph




Problem Description
 
In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are notadjacent in G. 

Now you are given an undirected graph G of N nodes and M bidirectional edges of unit length. Consider the complement of G, i.e., H. For a given vertex S on H, you are required to compute the shortest distances from S to all N−1 other vertices.
 

Input

 

There are multiple test cases. The first line of input is an integer T(1≤T<35) denoting the number of test cases. For each test case, the first line contains two integers N(2≤N≤200000) and M(0≤M≤20000). The following M lines each contains two distinct integers u,v(1≤u,v≤N) denoting an edge. And S (1≤S≤N) is given on the last line.
 

 

Output
 
For each of T test cases, print a single line consisting of N−1 space separated integers, denoting shortest distances of the remaining N−1 vertices from S (if a vertex cannot be reached from S, output ``-1" (without quotes) instead) in ascending order of vertex number.
 

 

Sample Input
 
12 01
 

 

Sample Output
 
1
 
 
题意：
　　
　　n 个点的无向完全图中删除 m 条边，问点 s 到其他点的最短路长度。
 
题解：
　　
　　技术分享
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<queue>#include<set>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair<int,int>#define MP make_pairtypedef long long LL;const long long INF = 1e18;const double Pi = acos(-1.0);const int N = 2e5+10, M = 1e2+11, mod = 1e9+7, inf = 2e9; int T,n,m,vis[N],head[N],t,d[N],ans[N]; struct ss{int to,next;}e[N * 2]; void add(int u,int v) {e[t].to=v;e[t].next=head[u];head[u]=t++;}int main() {        scanf("%d",&T);        while(T--) {            scanf("%d%d",&n,&m);            t =1;memset(head,0,sizeof(head));            for(int i = 1; i <= m; ++i) {                int a,b;                scanf("%d%d",&a,&b);                add(a,b);add(b,a);            }            memset(vis,0,sizeof(vis));            memset(d,-1,sizeof(d));            int S;            scanf("%d",&S);            queue<int > q;            q.push(S);vis[S] = 1;            d[S] = 0;            set<int > s;            for(int i = 1; i <= n; ++i) if(i != S) s.insert(i),vis[i] = 1;            while(!q.empty()) {                int k = q.front();                q.pop();                for(int i = head[k]; i; i =e[i].next) {                    int to = e[i].to;                    if(s.count(to)) {                        vis[to] = 0;                    }                }                for(set<int > ::iterator itt,it = s.begin();it != s.end(); ) {                    if(vis[*it])                    {                        d[*it] = d[k] + 1;                        q.push(*it);                        itt = it;                        itt++;                        s.erase(it);                        it = itt;                    } else {                        it++;                    }                }                for(set<int > ::iterator itt,it = s.begin();it != s.end(); it++) vis[*it] = 1;            }             int cnt = 0;            for(int i = 1; i <= n; ++i) {                    if(i != S)ans[++cnt] = d[i];            }            for(int i = 1; i < cnt; ++i) printf("%d ",ans[i]);            printf("%d\n",ans[cnt]);        }        return 0;}

 

HDU 5876 Sparse Graph BFS 最短路


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