标签：white continue transform growth after ima position int http
BaoBao and DreamGrid are playing the game Plants vs. Zombies. In the game, DreamGrid grows plants to defend his garden against BaoBao‘s zombies.
There are plants in DreamGrid‘s garden arranged in a line. From west to east, the plants are numbered from 1 to and the -th plant lies meters to the east of DreamGrid‘s house. The -th plant has a defense value of and a growth speed of . Initially, for all .
DreamGrid uses a robot to water the plants. The robot is in his house initially. In one step of watering, DreamGrid will choose a direction (east or west) and the robot moves exactly 1 meter along the direction. After moving, if the -th plant is at the robot‘s position, the robot will water the plant and will be added to . Because the water in the robot is limited, at most steps can be done.
The defense value of the garden is defined as . DreamGrid needs your help to maximize the garden‘s defense value and win the game.
Please note that:
There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:
The first line contains two integers and (, ), indicating the number of plants and the maximum number of steps the robot can take.
The second line contains integers (), where indicates the growth speed of the -th plant.
It‘s guaranteed that the sum of in all test cases will not exceed .
For each test case output one line containing one integer, indicating the maximum defense value of the garden DreamGrid can get.
2 4 8 3 2 6 6 3 9 10 10 1
6 4
In the explanation below, ‘E‘ indicates that the robot moves exactly 1 meter to the east from his current position, and ‘W‘ indicates that the robot moves exactly 1 meter to the west from his current position.
For the first test case, a candidate direction sequence is {E, E, W, E, E, W, E, E}, so that we have after the watering.
For the second test case, a candidate direction sequence is {E, E, E, E, W, E, W, E, W}, so that we have after the watering.
Author: WANG, Yucheng; CHEN, Shihan
Source: The
2018 ACM-ICPC Asia Qingdao Regional Contest
【题意】
从左到右依次是1个房子+n个植物，从房子出发给n个植物浇水，每一次可以往左可以往右，起始时植物的防御值都为0，每一次在i位置浇一次水，防御增加d[i]，每一次必须走，不能呆在原地，可以走出去，定义n个植物的防御力为min(ai),1<=i<=n,问怎样走使得这些植物防御力最小值最大
【分析】
l=0,r=1e12 二分可以的最小价值的最大值。以第一个样例为准，a[1]=3 a[2]=2 a[3]=6 a[4]=6 如果最小值为7，那么位置1就需要走3次，即走到2，走回1，走到2，走回1。所以当1位置被走了3次后，2位置其实已经被走过了2次。以此类推，用step记录走的步数，先算出在某个二分的值下每个位置最少需要走几步，然后遍历一遍，看看真实走的步数是否比m小，如果小，说明当前二分的值偏小，否则偏大
【代码】
#include<cstdio>
using namespace std;
typedef long long ll;
inline ll read(){
register char ch=getchar();register ll x=0;
for(;ch<‘0‘||ch>‘9‘;ch=getchar());
for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar()) x=(x<<3)+(x<<1)+ch-‘0‘;
return x;
}
const int N=1e5+5;
int T,n;ll m,a[N],c[N],d[N];
inline bool check(ll now){
if(!now) return 1;
ll tot=m;
for(int i=1;i<=n;i++) c[i]=(now-1)/a[i]+1,d[i]=0;
//简洁化if(now%a[i]) c[i]=now/a[i]+1;else c[i]=now/a[i];
for(int i=1;i<=n;i++){
if(i==n&&c[i]<=d[i]) return 1;
if(tot<=0) return 0;
d[i]++,tot--;
if(c[i]<=d[i]) continue;
ll need=c[i]-d[i];
if((need<<1)>tot) return 0;
tot-=need<<1;//因为想让当前位置满足次数的话，必定是一来一回，需要两倍的步数
d[i+1]+=need;//d[i]对d[i+1]的贡献
}
return 1;
}
int main(){
for(T=read();T--;){
n=read();m=read();
for(int i=1;i<=n;i++) a[i]=read();
ll l=0,r=1e12,ans=0;
while(l<=r){
ll mid=l+r>>1;
if(check(mid)){
ans=mid;
l=mid+1;
}
else{
r=mid-1;
}
}
printf("%lld\n",ans);
}
return 0;
}2018 青岛ICPC区域赛E ZOJ 4062 Plants vs. Zombie(二分答案)
标签：white continue transform growth after ima position int http
原文地址：https://www.cnblogs.com/shenben/p/10312053.html
