上海高校金马五校赛 J

题目大意： 给你n个数字， 定义不连贯值为， max(abs(a[ i ] - b[ i ])) ，现在让你把m个新的数字插入n + 1 个空位中，使得不连贯值最小。

思路：二分不连贯值， 每次进行二分图匹配， 注意进行二分图匹配的时候需要加入虚拟的点，因为这n + 1个空位中有些点是必须加数字的。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>
using namespace std;

const int N=400+7;
const int M=100+7;
const int inf=0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3f;
const int mod=1e9 + 9;

int n, m, a[N], b[N],cx[N], cy[N], vis[N];
vector<int> edge[N];

bool path(int u)
{
    for(int v : edge[u])
    {
        if(!vis[v])
        {
            vis[v] = 1;
            if(cy[v] == -1 || path(cy[v]))
            {
                cx[u] = v;
                cy[v] = u;
                return true;
            }
        }
    }
    return false;
}

bool maxmatch()
{
    int res=0;
    memset(cx, -1, sizeof(cx));
    memset(cy, -1, sizeof(cy));
    for(int i = 1; i <= 2 * n + 2; i++)
    {
        if(cx[i] == -1)
        {
            memset(vis, 0, sizeof(vis));
            if(path(i)) continue;
            else return false;
        }
    }
    return true;
 }

bool check(int x) {
    for(int i = 1; i <= 2 * n + 2; i++)
        edge[i].clear();
    for(int i = 1; i <= n + 1; i++) {
        if(i == 1) {
            for(int j = 1; j <= m; j++) {
                if(abs(b[j] - a[1]) <= x) {
                    edge[i].push_back(n + 1 + j);
                    edge[n + 1 + j].push_back(i);
                }
            }
            for(int j = m + 1; j <= n + 1; j++) {
                edge[i].push_back(n + 1 + j);
                edge[n + 1 + j].push_back(i);
            }
        } else if(i == n + 1) {
            for(int j = 1; j <= m; j++) {
                if(abs(b[j] - a[n]) <= x) {
                    edge[i].push_back(n + 1 + j);
                    edge[n + 1 + j].push_back(i);
                }
            }
            for(int j = m + 1; j <= n + 1; j++) {
                edge[i].push_back(n + 1 + j);
                edge[n + 1 + j].push_back(i);
            }
        } else {
            for(int j = 1; j <= m; j++) {
                if((abs(b[j] - a[i]) <= x) && abs(b[j] - a[i - 1]) <= x) {
                    edge[i].push_back(n + 1 + j);
                    edge[n + 1 + j].push_back(i);
                }
            }
            if(abs(a[i] - a[i - 1]) <= x) {
                for(int j = m + 1; j <= n + 1; j++) {
                    edge[i].push_back(n + 1 + j);
                    edge[n + 1 + j].push_back(i);
                }
            }
        }
    }
    return maxmatch();
}
int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for(int i = 1; i <= m; i++)
            scanf("%d", &b[i]);

        int l = 0, r = 1e9, ans = -1;

        while(l <= r) {
            int mid = (l + r) >> 1;
            if(check(mid)) {
                ans = mid;
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}
/***************
****************/