Problem Description
Soda has a bipartite graph with n vertices
and m undirected
edges. Now he wants to make the graph become a complete bipartite graph with most edges by adding some extra edges. Soda needs you to tell him the maximum number of edges he can add.
Note: There must be at most one edge between any pair of vertices both in the new graph and old graph.
Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤100),
indicating the number of test cases. For each test case:
The first line contains two integers n and m, (2≤n≤10000,0≤m≤100000).
Each of the next m lines
contains two integer u,v (1≤u,v≤n,v≠u) which
means there‘s an undirected edge between vertex u and
vertex v.
There‘s at most one edge between any pair of vertices. Most test cases are small.
Output
For each test case, output the maximum number of edges Soda can add.
Sample Input
2 4 2 1 2 2 3 4 4 1 2 1 4 2 3 3 4
Sample Output
2 0
Source
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大致题意：
有n个点，m条边的二分图（可能不连通），问最多还能加多少条边变成完全二分图
思路：
显然每一连通块，都染成两种颜色，最后要尽量使两种颜色总数相同解才最优
显然有两种决策，不是染白就是染黑，01背包
dp[i][val]表示前i个连通块能染成同一色点数<=val的最大值
显然dp[scc][all/2]是最优解
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
#define X first
#define Y second
typedef pair<ll,ll> pii;
const int N = 10000+100;
const int M = 100000+1000;
struct Edge{
int v,nxt;
Edge(int v = 0,int nxt = 0):v(v),nxt(nxt){}
}es[M*2];
int n,m;
int ecnt;
int head[N];
inline void add_edge(int u,int v){
es[ecnt] = Edge(v,head[u]);
head[u] = ecnt++;
es[ecnt] = Edge(u,head[v]);
head[v] = ecnt++;
}
int col[N];
int cnt[N][2];
int top;
int sum = 0;
void dfs(int u,int fa){
col[u] = !col[fa];
cnt[top][col[u]]++;
for(int i = head[u];~i;i = es[i].nxt){
int v = es[i].v;
if(v == fa || col[v] != -1) continue;
dfs(v,u);
}
}
void ini(){
REP(i,n) head[i] = col[i] = -1,cnt[i][0] = cnt[i][1] = 0;
col[0] = top = sum = ecnt = 0;
}
int dp[2][N];
int main(){
int T;
cin>>T;
while(T--){
scanf("%d%d",&n,&m);
ini();
REP(i,m){
int u,v;
scanf("%d%d",&u,&v);
add_edge(u,v);
}
for(int i = n; i>= 1;i--){
if(col[i] != -1) continue;
top++;
dfs(i,0);
if(cnt[top][0] == 0 || cnt[top][1] == 0) {
cnt[top][0] = cnt[top][1] = 0;
top--;
}
else {
sum += cnt[top][0],sum += cnt[top][1];
}
}
int nd = n-sum;
for(int i = 0;i <= sum/2;i++) dp[0][i] = 0;
REP(i,top){
for(int j = 0; j <= sum/2; j++){
dp[i&1][j] = -1;
if(j-cnt[i][0] >= 0 && dp[(i-1)&1][j-cnt[i][0]] != -1) dp[i&1][j] = dp[(i-1)&1][j-cnt[i][0]]+cnt[i][0];
if(j-cnt[i][1] >= 0 && dp[(i-1)&1][j-cnt[i][1]] != -1) {
dp[i&1][j] = max(dp[(i-1)&1][j-cnt[i][1]]+cnt[i][1],dp[i&1][j]);
}
}
int minn,maxx = sum-dp[top&1][sum/2];
int t = min(nd,maxx-dp[top&1][sum/2]);
minn = dp[top&1][sum/2]+t;
nd -= t;
if(nd) minn += nd/2, maxx += nd/2 + (nd&1);
printf("%d\n",minn*maxx-m);
}
}
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