UVa Live 4725 – Airport 二分，动态规划，细节 难度: 1

题目

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2726

题意

飞机，一个起飞的跑道，两个降落的跑道。每个时刻首先两个跑道降落一些飞机，然后再飞走一架飞机，最后所有还停留着的飞机按照0开始编号，问如何安排能使序号最小

思路

明显，使用二分法枚举答案

感想：

由于题目描述，不太能理解到底是如何起飞，降落，序号又是何时进行统计的，所以在最后的序号上是否要多取1卡了很久

代码

#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <tuple>
#define LOCAL_DEBUG
using namespace std;
typedef pair<int, int> MyPos;
const int MAXN = ;
int a[MAXN];
int b[MAXN];
int sumA[MAXN];
int sumB[MAXN];
int limitedA[MAXN];
int limitedB[MAXN];
int limitedSum[MAXN];
int n;
bool check(int mx) {
    int costA = , costB = ;
    for (int i = ; i <= n; i++) {
        costA = max(costA, sumA[i] - mx);
        costB = max(costB, sumB[i] - mx);
        if (costA > limitedA[i] || costB > limitedB[i] || costA + costB > limitedSum[i])return false;
    }
    return true;
}int main() {
#ifdef LOCAL_DEBUG
    freopen("C:\\Users\\Iris\\source\\repos\\ACM\\ACM\\input.txt", "r", stdin);
    freopen("C:\\Users\\Iris\\source\\repos\\ACM\\ACM\\output.txt", "w", stdout);
#endif // LOCAL_DEBUG
    int T;
    cin >> T;
    for (int ti = ; ti <= T && cin >> n; ti++) {
        for (int i = ; i <= n; i++) {
            cin >> a[i] >> b[i];
        }
        //if (ti < 258)continue;
        for (int i = ; i <= n; i++) {
            sumA[i] = sumA[i - ] + a[i];
            sumB[i] = sumB[i - ] + b[i];
            bool addA = sumA[i - ] > limitedA[i - ];
            bool addB = sumB[i - ] > limitedB[i - ];
            limitedA[i] = limitedA[i - ] + (addA ?  : );
            limitedB[i] = limitedB[i - ] + (addB ?  : );
            limitedSum[i] = limitedSum[i - ] + ((limitedA[i] + limitedB[i] > limitedSum[i - ]) ?  : );
        }
        int l = , r = n * ;
        while (l < r) {
            int mid = (l + r) >> ;
            if (mid == l)break;
            if (check(mid)) {
                r = mid;
            }
            else {
                l = mid;
            }
        }
        cout<<l<<endl;
    }
    return ;
}