hdu 5876 暴力

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Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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 not adjacent 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
1
2 0
1
 
Sample Output
1
 

题意：

n个点m条边的图，给定一个点S，要求输出再补图上，S到所有点的最短路径

思路：

找出独立的点X(没有边连)，那么S到X的距离为1，到其他的所有在图上的点距离为2

如果不存在独立的点，那么图很小，二维数组存边跑dijkstra

代码：

#include <iostream>#include <vector>#include <cmath>using namespace std;int vis[200100];int n,m,S,X;int anw[200100];int u[20010],v[20010];vector <int> V;int a[10010][10010];int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++) vis[i]=0;        for(int i=1;i<=n;i++) anw[i]=0;        V.clear();        for(int i=1;i<=m;i++) scanf("%d%d",&u[i],&v[i]);        scanf("%d",&S);        for(int i=1;i<=m;i++){            vis[u[i]]=vis[v[i]]=1;            if(u[i]==S) V.push_back(v[i]);            else if(v[i]==S) V.push_back(u[i]);        }        X=-1;        for(int i=1;i<=n;i++){            if(i!=S && !vis[i]) {                X=i;                break;            }        }        if(X!=-1){            int cnt=V.size();            for(int i=0;i<cnt;i++) anw[V[i]]=2;            for(int i=1;i<=n;i++) {                if(!anw[i]) anw[i]=1;            }            anw[S]=0;        } else {            for(int i=1;i<=n;i++){                for(int j=1;j<=n;j++){                    if(i==j) a[i][j]=0;                    else a[i][j]=a[j][i]=1;                }            }            for(int i=1;i<=m;i++){                a[u[i]][v[i]]=1e9;                a[v[i]][u[i]]=1e9;            }            for(int i=1;i<=n;i++) vis[i]=0;            for(int i=1;i<=n;i++) anw[i]=1e9;            anw[S]=0;            int flag=0;            for(int i=1;i<n;i++){                int f=-1,mn=1e9;                for(int j=1;j<=n;j++){                    if(!vis[j] && anw[j]<mn){                        f=j;                        mn=anw[j];                    }                }                if(f==-1){                    flag=1;                    break;                }                vis[f]=1;                for(int j=1;j<=n;j++){                    if(!vis[j] && anw[f]+a[f][j]<anw[j]){                        anw[j]=anw[f]+a[f][j];                    }                }            }            if(flag){                printf("-1\n");                continue;            }        }        if(S==1) {            for(int i=2;i<n;i++) printf("%d ",anw[i]);            printf("%d\n",anw[n]);        } else {            printf("%d",anw[1]);            for(int i=2;i<=n;i++){                if(anw[i]) printf(" %d",anw[i]);            }            printf("\n");        }    }    return 0;}/*104 31 22 33 42*/