HNUCM 1347 wjw的排隊問題 (二分 思維)

題目鏈接:http://acm.hnucm.edu.cn/JudgeOnline/problem.php?id=1347

二分找最小距離值，讓每個小團體移向兩邊的距離相等，就能得到最小值

判斷這個值v是否符合條件：

對於所有的n,使得b[i](b[i]爲第i個團體的位置) + v - 起始點 + 1 >= a[i](a[i]爲第i個團體人數)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#define rep(i, s, e) for(int i = s; i < e; ++i)
#define P pair<int, int>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
static const int N = 35;
static const int MAX_N = 1e6 + 5;
static const ll Mod = 1e9 + 7;
int a[MAX_N], b[MAX_N];
int n;
bool judge(int v){
    int d = -INF;
    for(int i = 0; i < n; ++i){
        d = max(d, b[i] - v);
        if(b[i] + v - d + 1 < a[i]) return false;
        d += a[i];
    }
    return true;
}
void solve(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) scanf("%d", &a[i]);
    for(int i = 0; i < n; ++i) scanf("%d", &b[i]);
    int l = 1, r = INF;
    while(l <= r){
        int mid = (l + r) >> 1;
        if(judge(mid)) r = mid - 1;
        else l = mid + 1;
    }
    printf("%d\n", l);
}
int main(){
    solve();
    return 0;
}