版权声明：From: http://blog.csdn.net/tju_tahara https://blog.csdn.net/TJU_Tahara/article/details/52549415
Sparse Graph
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1574 Accepted Submission(s): 553
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.
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
Source
Recommend
题意：
图的补图的某一点到所有点的最短路，分别输出。
这次比赛真的是水过去了，状态非常差。这道题用裸BFS来解最短路也是第一次遇到。题做多了脑子就死了。比较裸的bfs，手懒脑子又懒的直接用map储存原来有的边，假如map[a][b]为0,那么补图就为1, 遍历所有可到达的顶点即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <map>
#define MAXV 200010
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
map<int, bool> ma[MAXV];
int dist[MAXV];
bool oc[MAXV];
void clear(int v){
v += 3;
memset(oc, 0, sizeof(oc));
for(int i = 1; i < v; i++){
ma[i].clear();
dist[i] = INF;
}
}
void bfs(int s, int v){
int i, now, nxt;
queue<int> que1;
queue<int> que3;
queue<int> que2;
for(i = 1; i <= v; i++)
que1.push(i);
que2.push(s);
oc[s] = 1;
dist[s] = 0;
while((!que1.empty()) && (!que2.empty())){
now = que2.front();
que2.pop();
while(!que1.empty()){
nxt = que1.front();
que1.pop();
if(now == nxt) continue;
if(ma[now][nxt]){
que3.push(nxt); continue;
}
if(dist[nxt] > dist[now] + 1)
dist[nxt] = dist[now] + 1;
if(!oc[nxt]){
que2.push(nxt);
oc[nxt] = 1;
}
}
while(!que3.empty()){
nxt = que3.front();
que3.pop();
que1.push(nxt);
}
}
}
void print(int v){
int i, cnt;
cnt = 2; i = 1;
while(cnt < v){
if(dist[i] == 0){
++i; continue;
}
else if(dist[i] == INF){
printf("-1 "); ++i; ++cnt;
}
else{
printf("%d ", dist[i]);
++i; ++cnt;
}
}
while(cnt == v){
if(dist[i] == 0){
++i; continue;
}
else if(dist[i] == INF){
printf("-1\n"); ++i; cnt++;
}
else{
printf("%d\n", dist[i]);
++i; ++cnt;
}
}
}
int main()
{
int casen, v, e, i, s, t;
scanf("%d", &casen);
while(casen--){
scanf("%d%d", &v, &e);
clear(v);
for(i = 0; i < e; i++){
scanf("%d%d", &s, &t);
ma[t][s] = ma[s][t] = 1;
}
scanf("%d", &s);
bfs(s, v);
print(v);
}
return 0;
}