ZOJ4062 Plants vs. Zombies（二分+贪心）

题目链接：传送门

题目大意：

　　有n棵植物依次放在1-n，机器人从0出发浇水，每棵植物被浇水时d_i += a_i，求浇m次水后min｛d_i|1 ≤ i ≤ n｝的最大值。（浇水时必须往左或往右走一步，落脚点为浇水点）

思路：

　　若已知答案mid（滑稽脸）：可以贪心地从左到右浇水，反复给d_i和d_i+1浇水，直到当前的d_i ≥ mid。。。O(n)。。。既然可以O(n)验证答案，那就开心地二分掉了。

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int MAX_N = 1e5 + ;

int N;
ll M;
ll a[MAX_N];
ll d[MAX_N];

bool judge(ll mid) {
    for (int i = ; i <= N; i++) d[i] = ;
    ll cnt = , jumped = ;
    for (int i = ; i <= N; i++) {
        if (d[i] >= mid) {
            jumped++;
            continue;
        }
        cnt += jumped; jumped = ;
        ll mul = (mid-d[i])/a[i] + ((mid-d[i])%a[i] > );
        cnt += mul*-;
        d[i] += mul * a[i];
        if (i+ <= N)
            d[i+] += (mul-) * a[i+];
    }
    return cnt <= M;
}

int main()
{
    int T;
    cin >> T;
    while (T--) {
        scanf("%d%lld", &N, &M);
        ll _max = ;
        for (int i = ; i <= N; i++) {
            scanf("%lld", a+i);
            _max = max(_max, a[i]);
        }

        ll l = , r = _max*M;
        ll ans = ;
        while (l <= r) {
            ll mid = (l+r) >> ;
            if (judge(mid)) {
                ans = max(ans, mid);
                l = mid+;
            }
            else
                r = mid-;
        }
        cout << ans << endl;
    }
    return ;
}
/*
2
4 8
3 2 6 6
3 9
10 10 1
*/