UVa 1450 Ariport

方法：二分

二分答案，检查的时候，记录下当前round之前能飞走的飞机数。如果一个方向为空，另一个方向有飞机，那就让另一个方向飞走一架飞机。如果两边边飞机数超过二分的限制的部分超过可以飞走的飞机数，那就返回false。注意，设置不需要更新当前飞机的数量。

code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

const int maxn = 5e3+5;
int a[2][maxn];
int n;
bool valid(int mid)
{
    int r[2]={0,0}, moves = 0;
    repn(i, n)
    {
        rep(j, 2) r[j] += a[j][i];
        int cut = 0;
        rep(j, 2) if (r[j] > mid+1)
        {
            cut += r[j]-mid-1;
        }
        if (cut > moves) return false;
        if (!r[0] && r[1]) r[1] -= 1;
        else if (r[0] && !r[1]) r[0] -= 1;
        else if (r[0] && r[1] && r[0] + r[1] > moves)
            ++moves;
    }
    return true;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;  cin >> T;
    repn(kase, T)
    {
        cin >> n;
        repn(i, n) rep(j, 2) cin >> a[j][i];
        int left = 0, right = 1e5+5, mid, ans=0;
        while (left <= right)
        {
            mid = (left+right)>>1;
            if (valid(mid))
            {
                ans = mid;
                right = mid-1;
            }
            else
                left = mid+1;
        }
        cout << ans << newline;
    }
}
/*
 3
 1 
 1 1
 3
 3 2
 0 3
 2 0
 6 
 0 1
 1 1
 1 2
 1 1
 1 1
 6 0
*/