poj 3253 Fence Repair 贪心
2015-12-03 15:43
423 查看
Fence Repair
Description
Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤
50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made;
you should ignore it, too.
FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.
Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.
Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will
result in different charges since the resulting intermediate planks are of different lengths.
Input
Line 1: One integer N, the number of planks
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts
Sample Input
Sample Output
Hint
He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into
16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).
Source
挑战上的,,按挑战的方法过的,比较暴力,注意ans要用long long 因为数比较大而且还要多次计算,int 会爆
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[20005];
typedef long long ll;
int swap(int &a,int &b)
{
int temp=a;
a=b;
b=temp;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
ll ans=0;
int min1=1,min2=2,t;
sort(a+1,a+n+1);
while(n>1)
{
min1=1,min2=2;
if(a[min1]>a[min2])
swap(min1,min2);
for(int i=3;i<=n;i++)
if(a[i]<a[min1])
{
min2=min1;
min1=i;
}
else if(a[i]<a[min2])
{
min2=i;
} //找出最小的数和次小的数
t=a[min1]+a[min2];
ans+=t;
if(min1==n) swap(min1,min2);
a[min1]=t;
a[min2]=a
; //补足空位
n--;
}
printf("%lld\n",ans);
}
return 0;
}
优先队列解决:
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
struct Node{
int l;
bool operator <(Node b) const{
return l>b.l;
};
}node[20005];
int main()
{
int n;
while(~scanf("%d",&n))
{
priority_queue<Node> q;
for(int i=1;i<=n;i++)
{
scanf("%d",&node[i].l);
q.push(node[i]);
}
int min1,min2;
long long ans=0;
Node temp;
while(q.size()>1)
{
min1=q.top().l;
q.pop();
min2=q.top().l;
q.pop();
temp.l=min1+min2;
ans+=temp.l;
q.push(temp);
}
printf("%lld\n",ans);
}
return 0;
}
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 34406 | Accepted: 11071 |
Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤
50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made;
you should ignore it, too.
FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.
Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.
Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will
result in different charges since the resulting intermediate planks are of different lengths.
Input
Line 1: One integer N, the number of planks
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts
Sample Input
3 8 5 8
Sample Output
34
Hint
He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into
16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).
Source
挑战上的,,按挑战的方法过的,比较暴力,注意ans要用long long 因为数比较大而且还要多次计算,int 会爆
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[20005];
typedef long long ll;
int swap(int &a,int &b)
{
int temp=a;
a=b;
b=temp;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
ll ans=0;
int min1=1,min2=2,t;
sort(a+1,a+n+1);
while(n>1)
{
min1=1,min2=2;
if(a[min1]>a[min2])
swap(min1,min2);
for(int i=3;i<=n;i++)
if(a[i]<a[min1])
{
min2=min1;
min1=i;
}
else if(a[i]<a[min2])
{
min2=i;
} //找出最小的数和次小的数
t=a[min1]+a[min2];
ans+=t;
if(min1==n) swap(min1,min2);
a[min1]=t;
a[min2]=a
; //补足空位
n--;
}
printf("%lld\n",ans);
}
return 0;
}
优先队列解决:
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
struct Node{
int l;
bool operator <(Node b) const{
return l>b.l;
};
}node[20005];
int main()
{
int n;
while(~scanf("%d",&n))
{
priority_queue<Node> q;
for(int i=1;i<=n;i++)
{
scanf("%d",&node[i].l);
q.push(node[i]);
}
int min1,min2;
long long ans=0;
Node temp;
while(q.size()>1)
{
min1=q.top().l;
q.pop();
min2=q.top().l;
q.pop();
temp.l=min1+min2;
ans+=temp.l;
q.push(temp);
}
printf("%lld\n",ans);
}
return 0;
}
相关文章推荐
- React Native 开发环境部署
- 远程管理HTML
- jQuery可悬停控制图片轮播
- uwsgi + nigix + django的样式展示
- js控制浏览器后退
- escape()、encodeURI()、encodeURIComponent()区别详解
- 在流程图中求支配点的一种快速算法+[CodeChef FEB14]Graph Challenge解题报告(求半支配点)
- css书写规范class和id常见的命名
- jQuery移动页面开发中的触摸事件与虚拟鼠标事件简介
- JS根据动态生成的字符串,验证是否存在对应function并执行
- JS正则获取参数值
- 使用 后退键/history.back()出现"警告: 网页已过期的解决办法"
- CSS DIV 居中
- Java NIO系列教程(三) Buffer
- xStream完美转换XML、JSON
- 基于webpack的前端工程化开发解决方案探索(一):动态生成HTML
- css定位的top定位
- ajax请求中传递的参数中如果含有特殊字符怎么处理?
- html5页面中拨打电话和发短信方式
- html高度随分辨度改变而改变