Beijing Guards
Time limit: 3.000 seconds
The guard had to keep an eye on his section of the wall all day, so he had to stay in the tower. This is a very boring job, thus it is important to keep the guards motivated. The best way to motivate a guard is to give him lots of awards. There are several different types of awards that can be given: the Distinguished Service Award, the Nicest Uniform Award, the Master Guard Award, the Superior Eyesight Award, etc. The Central Department of City Guards determined how many awards have to be given to each of the guards. An award can be given to more than one guard. However, you have to pay attention to one thing: you should not give the same award to two neighbors, since a guard cannot be proud of his award if his neighbor already has this award. The task is to write a program that determines how many different types of awards are required to keep all the guards motivated
Input
The input contains several blocks of test eases. Each case begins with a line containing a single integer ln100000, the number of guard towers. The next n lines correspond to the n guards: each line contains an integer, the number of awards the guard requires. Each guard requires at least 1, and at most l00000 awards. Guard i and i + 1 are neighbors, they cannot receive the same award. The first guard and the last guard are also neighbors.
The input is terminated by a block with n = 0.
Output
For each test case, you have to output a line containing a single integer, the minimum number x of award types that allows us to motivate the guards. That is, if we have x types of awards, then we can give as many awards to each guard as he requires, and we can do it in such a way that the same type of award is not given to neighboring guards. A guard can receive only one award from each type.
Sample Input
3
4
2
2
5
2
2
2
2
2
5
1
1
1
1
1
0
Sample Output
8
5
3
题目大意：n个人围成圈，第i个人想要ri种礼物，相邻两个人不能有相同的礼物，问至少需要多少种礼物才可满足所有人的要求。
思路：
人数为偶数的情况就是相邻的两个ri想加的最大值p(很好理解),分配策略为，编号i为奇数的人分配礼物1~ri的礼物。而编号i为偶数的人分配礼物编号为p-ri+1~p。所以一定能避免冲突。
人数为奇数的情况：二分答案，若有p种礼物，第一个人把礼物分成两部分。前半部分为1~r1的礼物。后半部分为p-r1+1~p的礼物。最优策略为。编号为奇数的人优先取后半部分的礼物而偶数的人优先取前半部分的礼物。这样就有效减少了冲突率。如果最后一个人和第一个人礼物没有冲突那么p可行。
详细见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=100010;
//typedef __int64 ll;
int aw[maxn],le[maxn],ri[maxn];
int n;
bool isok(int p)
{
int x,y,i;
x=aw[1],y=p-aw[1];
le[1]=x,ri[1]=0;
for(i=2;i<=n;i++)
{
if(i&1)
{
ri[i]=min(aw[i],y-ri[i-1]);
le[i]=aw[i]-ri[i];
}
else
{
le[i]=min(aw[i],x-le[i-1]);
ri[i]=aw[i]-le[i];
}
}
return le[n]==0;
}
int main()
{
int i,ans,low,hi,mid;
while(scanf("%d",&n),n)
{
for(i=1;i<=n;i++)
scanf("%d",&aw[i]);
if(n==1)
{
printf("%d\n",aw[1]);
continue;
}
low=hi=0;
for(i=2;i<=n;i++)
low=max(low,aw[i-1]+aw[i]);
low=max(aw[1]+aw[n],low);
if(n&1)
{
for(i=1;i<=n;i++)
hi=max(hi,3*aw[i]);
while(low<=hi)
{
mid=(low+hi)>>1;
if(isok(mid))
{
ans=mid;
hi=mid-1;
}
else
low=mid+1;
}
}
else
ans=low;
printf("%d\n",ans);
}
return 0;
}