给出两个长度为n的序列。这两个序列的数字可以连边当且仅当它们不同，权值为它们的绝对值，求出这个二分图的最小权值完全匹配。没有输出-1.
n<=1e5.用KM会TLE+MLE.
如果连边没有限制的话，将两个序列排序一下显然就得到最优解了。
考虑限制。则需要将排序后一些项交换。可以证明最优解最多交换距离为3。因为DP一下就可以了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-9 # define MOD 1000000000 # define INF 0x7ffffffffffll # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=100005; //Code begin... int a[N], b[N]; LL dp[N]; LL get(int a, int b){return a==b?INF:abs(a-b);} int main () { int n; scanf("%d",&n); FOR(i,1,n) scanf("%d%d",a+i,b+i); sort(a+1,a+n+1); sort(b+1,b+n+1); dp[1]=get(a[1],b[1]); if (n>=2) dp[2]=min(dp[1]+get(a[2],b[2]),get(a[1],b[2])+get(a[2],b[1])); FOR(i,3,n) { dp[i]=dp[i-1]+get(a[i],b[i]); dp[i]=min(dp[i],dp[i-2]+get(a[i],b[i-1])+get(a[i-1],b[i])); dp[i]=min(dp[i],dp[i-3]+get(a[i-2],b[i-1])+get(a[i-1],b[i])+get(a[i],b[i-2])); dp[i]=min(dp[i],dp[i-3]+get(a[i-2],b[i])+get(a[i-1],b[i-2])+get(a[i],b[i-1])); dp[i]=min(dp[i],dp[i-3]+get(a[i-2],b[i])+get(a[i-1],b[i-1])+get(a[i],b[i-2])); } if (dp[n]>=INF) puts("-1"); else printf("%lld ",dp[n]); return 0; }