Sparse Graph(HDU 5876)

大连网络赛的倒数第二题，比赛时想的太多了，没有A掉，挺可惜的！以下是题目

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1495    Accepted Submission(s): 525

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

12 01

 

Sample Output

1

题目的意思就是给你一个图与及一个起点，再根据给出的图找出这个图完全图的补图，最后求出补图上除起点外每个点到起点的距离（以为边没有权值，距离为起点到该点的最小边数）。

解题思路：

首先储存一下刚开始时图的边，而这些边在补图上不会存在，如果没有储存的边都是补图上的边， 这样就可以做出了补图。然后在补图上找出距离从1到n-1的点（因为一个图上两点之间的最长距离为n-1），找不到就无法到达起点，距离用一个数组储存。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <queue>using namespace std;const int big = 2e5 + 7;const int INF = 0x7f7f7f7f;vector<int>z[big];int dis[big], n, m, k, record[big], num1, num2;int fuck[big], sum, mid[big], amout;bool vis[big];void initialization(){    for(int i = 1; i <= n; i ++)    {        dis[i] = -1;        z[i].clear();    }}bool judge(int key){    int temp = 0;    for(int i = 0, j = z[key].size(); i < j; i ++)        if(vis[z[key][i]]) temp ++;    return temp < num2;}void bfs(){    bool flag;    for(int i = 1, d; i < n; i ++)    {        flag = true;        sum = 0;        for(int j = 1; j <= num1; j ++)        {            d = record[j];            if(!judge(d)) continue;            flag = false;            fuck[++ sum] = d;            record[j] = 0;        }        if(flag) return; /// 此处是优化，因为当距离为i时，没有点能到达，距离大于i的点不存在！        for(int j = 1; j <= sum; j ++)        {            d = fuck[j];            vis[d] = true;            dis[d] = i;        }        amout = 0;        for(int j = 1; j <= num1; j ++)        {            if(!record[j]) continue;            mid[++ amout] = record[j];        }        for(int j = 1; j <= amout; j ++)            record[j] = mid[j];        num1 -= sum;        num2 += sum;    }}int main(){    int t, u, v;    scanf("%d", &t);    while(t --)    {        scanf("%d %d", &n, &m);        initialization();        for(int i = 1; i <= m; i ++)        {            scanf("%d %d", &u, &v);            z[u].push_back(v);            z[v].push_back(u);        }        scanf("%d", &k);        memset( vis, false, sizeof( vis));        num1 = 0;        for(int i = 1; i <= n; i ++)        {            if(i == k) continue;            record[++ num1] = i;        }        num2 = n - num1;        dis[k] = 0;        vis[k] = true;        bfs();        for(int i = 1, s = 0; i <= n; i ++)        {            if(!dis[i]) continue;            if(s ++) printf(" ");            printf("%d", dis[i]);        }        printf("\n");    }    return 0;}