Codeforces Round #250 (Div. 2) 437C The Child and Toy(脑洞贪心)
2015-08-31 20:34
525 查看
C. The Child and Toy
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of n parts and m ropes.
Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part i as vi.
The child spend vf1 + vf2 + ... + vfk energy
for removing part i where f1, f2, ..., fk are
the parts that are directly connected to the i-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all n parts.
Input
The first line contains two integers n and m (1 ≤ n ≤ 1000; 0 ≤ m ≤ 2000).
The second line contains n integers: v1, v2, ..., vn(0 ≤ vi ≤ 105).
Then followed m lines, each line contains two integers xi and yi,
representing a rope from part xi to
part yi(1 ≤ xi, yi ≤ n; xi ≠ yi).
Consider all the parts are numbered from 1 to n.
Output
Output the minimum total energy the child should spend to remove all n parts of the toy.
Sample test(s)
input
output
input
output
input
output
Note
One of the optimal sequence of actions in the first sample is:
First, remove part 3, cost of the action is 20.
Then, remove part 2, cost of the action is 10.
Next, remove part 4, cost of the action is 10.
At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts.
一个礼物由n部分及m条绳子组成,每次拆掉一个部分有代价v,问你最小代价是多少。
看似麻烦的题,往往代码都短,每次都选出小的加起来就好了。。
AC代码:
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
On Children's Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he can't wait to destroy the toy.
The toy consists of n parts and m ropes.
Each rope links two parts, but every pair of parts is linked by at most one rope. To split the toy, the child must remove all its parts. The child can remove a single part at a time, and each remove consume an energy. Let's define an energy value of part i as vi.
The child spend vf1 + vf2 + ... + vfk energy
for removing part i where f1, f2, ..., fk are
the parts that are directly connected to the i-th and haven't been removed.
Help the child to find out, what is the minimum total energy he should spend to remove all n parts.
Input
The first line contains two integers n and m (1 ≤ n ≤ 1000; 0 ≤ m ≤ 2000).
The second line contains n integers: v1, v2, ..., vn(0 ≤ vi ≤ 105).
Then followed m lines, each line contains two integers xi and yi,
representing a rope from part xi to
part yi(1 ≤ xi, yi ≤ n; xi ≠ yi).
Consider all the parts are numbered from 1 to n.
Output
Output the minimum total energy the child should spend to remove all n parts of the toy.
Sample test(s)
input
4 3 10 20 30 40 1 4 1 2 2 3
output
40
input
4 4 100 100 100 100 1 2 2 3 2 4 3 4
output
400
input
7 10
40 10 20 10 20 80 401 5
4 7
4 5
5 2
5 7
6 4
1 6
1 3
4 3
1 4
output
160
Note
One of the optimal sequence of actions in the first sample is:
First, remove part 3, cost of the action is 20.
Then, remove part 2, cost of the action is 10.
Next, remove part 4, cost of the action is 10.
At last, remove part 1, cost of the action is 0.
So the total energy the child paid is 20 + 10 + 10 + 0 = 40, which is the minimum.
In the second sample, the child will spend 400 no matter in what order he will remove the parts.
一个礼物由n部分及m条绳子组成,每次拆掉一个部分有代价v,问你最小代价是多少。
看似麻烦的题,往往代码都短,每次都选出小的加起来就好了。。
AC代码:
#include "iostream" #include "cstdio" #include "cstring" #include "algorithm" using namespace std; const int maxn = 1005; int n, m, v[maxn]; int main(int argc, char const *argv[]) { cin >> n >> m; for(int i = 1; i <= n; ++i) cin >> v[i]; int ans = 0, x, y; while(m--) { cin >> x >> y; ans += min(v[x], v[y]); } cout << ans << endl; return 0; }
相关文章推荐
- 动态增加文本框
- 排序算法(三)
- 用两个栈实现队列
- 【LeetCode】之Linked List Cycle
- 用自定义缓存区的方式实现文件的移动
- c++知识点总结(不时更新)
- POST请求的两种方式
- scala实现设计模式之命令模式
- Thinking in Java---线程通信+三种方式实现生产者消费者问题
- 欢迎使用CSDN-markdown编辑器
- MVC中的筛选器
- 论版本控制
- hdu 4902 Nice boat(线段树区间更新lazytag·单点更新)
- Spring整合JMS(三)——MessageConverter介绍
- HDU 4552 怪盗基德的挑战书
- [ 收集] 条件注释判断浏览器版本
- 数据结构中排序算法-归并排序(4)
- Switch
- 使用call_user_func_array()来回调执行函数与直接使用函数的区别是什么? 周梦康周梦康 839 2014年04月15日 提问 · 2014年04月15日 更新 关注 0 关注 收藏
- MVC模式