BZOJ 1237 配对(DP)

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\n",dp[n]);
    return 0;
}