UVa 1335 Beijing Guards (二分+贪心)

题意：n 个人成一个圈，每个人想要 ri 种不同的礼物，要求相邻两个人没有相同的，求最少需要多少礼物。

析：如果 n 是偶数，那么答案一定是相邻两个人的礼物总种数之和的最大值，那么如果是奇数，就没那么好做了，我们可以二分答案，

在每次判定时，我们可以有这样的贪心策略，第一个人 1 - r1，在后面的人中，编号为奇数的尽量往后取，编号为偶数的尽量往前取，

因为这样我们才能保证第 n 个人和第一个人尽量不相交。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
 
int a[maxn];
int l[maxn], r[maxn];
 
bool judge(int mid){
    l[1] = a[1], r[1] = 0;
    int x = a[1], y = mid - a[1];
    for(int i = 2; i <= n; ++i){
        if(i & 1){
            r[i] = min(a[i], y-r[i-1]);
            l[i] = a[i] - r[i];
            if(l[i] + l[i-1] > x)  return false;
        }
        else{
            l[i] = min(a[i], x-l[i-1]);
            r[i] = a[i] - l[i];
            if(r[i] + r[i-1] > y)  return false;
        }
    }
    return l[n] == 0;
}
 
int solve(){
    if(n & 1){
        int l = 1, r = 5e5;
        while(l < r){
            int mid = (r + l) >> 1;
            if(judge(mid)) r = mid;
            else  l = mid + 1;
        }
        return l;
    }
    int ans = 0;
    for(int i = 1; i <= n; ++i) ans = max(ans, a[i] + a[i+1]);
    return ans;
}
 
int main(){
    while(scanf("%d", &n) == 1 && n){
        for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
        if(1 == n){ printf("%d\n", a[1]);  continue; }
        a[n+1] = a[1];
        printf("%d\n", solve());
    }
    return 0;
}
/*
9
8
15
35
16
21
90
55
50
32
*/