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;
}