codeforces 437C The Child and Toy
1 second
256 megabytes
standard input
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.
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 the minimum total energy the child should spend to remove all n parts of the toy.
4 3
10 20 30 40
1 4
1 2
2 3
40
4 4
100 100 100 100
1 2
2 3
2 4
3 4
400
7 10
40 10 20 10 20 80 40
1 5
4 7
4 5
5 2
5 7
6 4
1 6
1 3
4 3
1 4
160
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.
翻译
在儿童节这一天,我们的小朋友收到了来自Delayyy的礼物:一个玩具。
只是。小朋友实在是太淘气了。他迫不及待的要拆掉这个玩具。
这个玩具由n个部分和m条绳子组成。每条绳子连接着两个部分,而且不论什么两个部分至多由一条绳子直接相连。为了拆掉这个玩具,小朋友必须移除它的全部n个部分。每次他仅仅能移除一个部分。而且移除须要能量。
每个部分都有一个能量值v[i],他移除第i部分所需的能量是v[f[1]]+v[f[2]]+...+v[f[k]],当中f[1],f[2],...,f[k]是与i直接相连(且还未被移除)的部分的编号。
请帮助他找到移除n个部分所需的最小总能量。
Input
第一行包括两个整数n,m(1<=n<=10000;0<=m<=2000)。第二行包括n个整数v[1],v[2],...,v[n](0<=v[i]<=10^5)。接下来有m个行,每一行有两个整数x[i]和y[i],表示一条连接x[i]和y[i]的绳子(1<=x[i],y[i]<=n; x[i]不等于y[i])。
Output
输出所需的最小能量。
题解
有没有感觉看了这么长的题目非常累。事实上代码就这么短……
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n,m,w[1005];
long long ans=0;
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&w[i]);
for(int i=1;i<=m;i++)
{
int x,y;
scanf("%d %d",&x,&y);
ans+=min(w[x],w[y]);
}
printf("%lld",ans);
return 0;
}
codeforces 437C The Child and Toy的更多相关文章
- Codeforces 437C The Child and Toy(贪心)
题目连接:Codeforces 437C The Child and Toy 贪心,每条绳子都是须要割断的,那就先割断最大值相应的那部分周围的绳子. #include <iostream> ...
- CF(437C)The Child and Toy(馋)
意甲冠军:给定一个无向图,每个小点右键.操作被拉动所有点逐一将去,直到一个点的其余部分,在连边和点拉远了点,在该点右点的其余的费用.寻找所需要的最低成本的运营完全成本. 解法:贪心的思想,每次将剩余点 ...
- Codeforces Round #250 (Div. 1) A. The Child and Toy 水题
A. The Child and Toy Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/438/ ...
- Codeforces Round #250 Div. 2(C.The Child and Toy)
题目例如以下: C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input ...
- Codeforces The Child and Toy
The Child and Toy time limit per test1 second On Children's Day, the child got a toy from Delayyy as ...
- Codeforces 437B The Child and Set
题目链接:Codeforces 437B The Child and Set 開始是想到了这样的情况,比方lowbit之后从大到小排序后有这么几个数,200.100,60.50.S = 210.那先选 ...
- cf437C The Child and Toy
C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces 437E The Child and Polygon(间隔DP)
题目链接:Codeforces 437E The Child and Polygon 题目大意:给出一个多边形,问说有多少种切割方法.将多边形切割为多个三角形. 解题思路:首先要理解向量叉积的性质,一 ...
- Codeforces 437A The Child and Homework
题目链接:Codeforces 437A The Child and Homework 少看了一个条件,最后被HACK掉到203名,要不然就冲到100多一点了==.. 做这个题收获最大的是英语,A t ...
随机推荐
- 我常用的iphone开发学习网站[原创]
引用地址:http://www.cnblogs.com/fuleying/archive/2011/08/13/2137032.html Google 翻译 Box2d 托德的Box2D的教程! Bo ...
- BZOJ 1627: [Usaco2007 Dec]穿越泥地( BFS )
BFS... --------------------------------------------------------------------------------------- #incl ...
- QComboBox 添加图片(自带addItem函数就有这个功能,从没有注意过)
方法: 使用 QComboxBox::addItem(QIcon, QString); 示例: 点击(此处)折叠或打开 QComboBox *combo_status = new QComboB ...
- QT VS检测内存泄漏
测试程序:http://download.csdn.net/detail/ajaxhe/4085447 vld-2.2.3: http://vld.codeplex.com/releases/view ...
- 我在北京找工作(二):java实现算法<1> 冒泡排序+直接选择排序
工作.工作.找工作.经过1个多星期的思想斗争还是决定了找JAVA方面的工作,因为好像能比PHP的工资高点.呵呵 :-) (其实我这是笑脸,什么QQ输入法,模拟表情都没有,忒不人性化了.) 言归正传, ...
- fedora audacious 不能播放音乐
采用命令sudo dnf search audacious 可以看到 audacious-plugins-freeworld-mp3.x86_64 : MP3 playback plugin for ...
- Java学习之IO字节流
字节流分为FileInputStream 和FileOutputStream package com.io; import java.io.File; import java.io.FileInput ...
- mmc一维下料测试
另一组数据, 长度 = 6000; 切割长度 = {1664, 1599, 1552, 1409, 1352, 802, 660}; 需求数量 = {32, 96, 160, 16, 384, 112 ...
- 误mlogc.c:32:23: error: curl/curl.h: No such file or directory
出现以下错误: mlogc.c:32:23: error: curl/curl.h: No such file or directory mlogc.c:1091: error: expected ' ...
- POJ 2104(K-th Number-区间第k大-主席树)
K-th Number Time Limit: 20000MS Memory Limit: 65536K Total Submissions: 31790 Accepted: 9838 Cas ...