UVA1450-Airport

题目链接

题意：有一个飞机场。有两条待飞跑到w和e。一条起飞跑道。每一时刻仅仅能起飞一架飞机，然后有w[i]和e[i]架飞机进入w和e跑道。飞机编号从0開始，问说怎样安排起飞能够使得飞机编号的最大值最小。

思路：仅仅要二分搜索，找到最小的答案就能够了。注意跑道上为0时。是没有飞机起飞的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 5005;

int t;
int A[MAXN], B[MAXN];

int judge(int n) {
    int cura = 0, curb = 0, num = 0;
    int a, b;
    for (int i = 0; i < t; i++) {
        cura += A[i];
        curb += B[i];
        a = max(cura - n, 0);
        b = max(curb - n, 0);
        if (a + b > num)
            return false;
        if (cura > 0 && curb > 0 && cura + curb > num)
            num++;
        else if (cura > 0 && curb == 0)
            cura--;
        else if (cura == 0 && curb > 0)
            curb--;
    }
    return true;
}

int main() {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d", &t);
        for (int i = 0; i < t; i++)
            scanf("%d%d", &A[i], &B[i]);

        int L = 1, R = MAXN * 20, mid;
        while (L < R) {
            mid = (R + L) / 2;
            if (judge(mid))
                R = mid;
            else
                L = mid + 1;
        }
        printf("%d\n", L - 1);
    }
    return 0;
}