1065 最小正子段和 二分答案 + 判定

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1065

我的思路比较笨，我是直接二分那个答案mid

然后进行一次O(nlogn)的判定，如果能找到一个区间的和比mid小的，（当然这个区间的和也是要大于0），那就return true

进行判断如下：

处理出前缀和dp[i]，对于每一个i

目标是：在前i - 1个中，查看是否有这样的一个x，使得，dp[i] - x    <=   mid，&& dp[i] - x >= 1

二分找x，把前i - 1个东西压去set中，然后二分找x >= dp[i] - mid的那个x，判定即可。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 50000 + 20;
LL dp[maxn];
int n;
set<LL>ss;
bool check(LL val) {
    if (dp[1] <= val && dp[1] >= 1) return true;
    ss.clear();
    ss.insert(0);
    ss.insert(dp[1]);
    LL mx = max(0LL, dp[1]);
    for (int i = 2; i <= n; ++i) {
        LL x = dp[i] - val;
        if (mx < x) {
            mx = max(mx, dp[i]);
            ss.insert(dp[i]);
            continue;
        }
        set<LL> :: iterator it = ss.lower_bound(x);
        if (dp[i] - *it >= 1) return true;
        mx = max(mx, dp[i]);
        ss.insert(dp[i]);
    }
    return false;
}
void work() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        int x;
        scanf("%d", &x);
        dp[i] = dp[i - 1] + x;
    }
//    for (int i = 1; i <= n; ++i) {
//        cout << dp[i] << " ";
//    }
//    cout << endl;
    LL be = 1, en = 1e18L;
//    cout << check(1) << endl;
    while (be <= en) {
        LL mid = (be + en) >> 1;
        if (check(mid)) {
            en = mid - 1;
        } else be = mid + 1;
    }
//    cout << be << endl;
    printf("%lld\n", be);
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}