CSU 1116 Kingdoms

传送门

Description

A kingdom has n cities numbered 1 to n, and some bidirectional roads connecting cities. The capital is always city 1.

After a war, all the roads of the kingdom are destroyed. The king wants to rebuild some of the roads to connect the cities, but unfortunately, the kingdom is running out of money. The total cost of rebuilding roads should not exceed K.

Given the list of m roads that can be rebuilt (other roads are severely damaged and cannot be rebuilt), the king decided to maximize the total population in the capital and all other cities that are connected (directly or indirectly) with the capital (we call it "accessible population"), can you help him?

Input

The first line of input contains a single integer T (T<=20), the number of test cases.

Each test case begins with three integers n(4<=n<=16), m(1<=m<=100) and K(1<=K<=100,000).

The second line contains n positive integers pi (1<=pi<=10,000), the population of each city.

Each of the following m lines contains three positive integers u, v, c (1<=u,v<=n, 1<=c<=1000), representing a destroyed road connecting city u and v, whose rebuilding cost is c.

Note that two cities can be directly connected by more than one road, but a road cannot directly connect a city and itself.

Output

For each test case, print the maximal accessible population.

Sample Input

2
4 6 6
500 400 300 200
1 2 4
1 3 3
1 4 2
4 3 5
2 4 6
3 2 7
4 6 5
500 400 300 200
1 2 4
1 3 3
1 4 2
4 3 5
2 4 6
3 2 7

Sample Output

1100
1000

--------------------------------------------------------

裸状压dp。

状压dp易写错的地方

1. for循环变量比较多，i, j, k 什么的容易和全局变量搞混，而且循环变量/循环体本身也容易写错。

2. 位运算操作符容易和对应的逻辑运算操作符搞混。

----------------------------------------------------------------

此题有坑---重边

----------------------------------------------------------------

#include <bits/stdc++.h>using namespace std;
int g[18][18], p[18];
const int N(1<<18);
int dp[N], co[N];
int T, n, m, k, u, v, c, ans;
inline int ones(int x){int res=0;for(int i=1; i<=n; i++){if(x&1<<i) res++;}return res;
}
inline void trans(int s, int i){for(int j=1; j<=n; j++){if(g[i][j]&&!(s&1<<j)&&co[s]+g[i][j]<=k){int t=s|1<<j;dp[t]=dp[s]+p[j];ans=max(ans, dp[t]);co[t]=co[t]?min(co[t], co[s]+g[i][j]):co[s]+g[i][j];}}
}
int main(){freopen("in", "r", stdin);for(cin>>T; T--;){cin>>n>>m>>k;for(int i=1; i<=n; i++) cin>>p[i];for(;m--;){cin>>u>>v>>c;if(!g[u][v]||g[u][v]>c) g[u][v]=g[v][u]=c;}dp[1<<1]=ans=p[1], co[1<<1]=0;for(int i=1; i<n; i++){ //iterate over onesfor(int j=1<<1, tot=1<<n+1; j<tot; j++){ //iterate over stateif(ones(j)==i&&dp[j]){for(int k=1; k<=n; k++){ //iterate over bitsif(j&1<<k){trans(j, k);}}}}}printf("%d\n", ans);memset(dp, 0, sizeof(dp));memset(co, 0, sizeof(co));memset(g, 0, sizeof(g));}
}

转载于:https://www.cnblogs.com/Patt/p/4730925.html

CSU 1116 Kingdoms相关推荐

qt5 linux 控制台乱码,qt5.12 解决显示中文乱码问题
Python之Mac上搭建集成开发环境首先下载一个东西: 找到下载地址:https://download.jetbrains.8686c.com/python/pycharm-professiona ...
CSU 1113 Updating a Dictionary（map容器应用）
题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1113 解题报告:输入两个字符串,第一个是原来的字典,第二个是新字典,字典中的元素的格式为 ...
csu 1554: SG Value 思维题
http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1554 这题在比赛的时候居然没想出来,然后发现居然是做过的题目的变种!!!! 先不考虑插入操作, ...
模拟 CSU 1562 Fun House
题目传送门 1 /* 2 题意:光线从 '*' 发射,遇到 '/' 或 '\' 进行反射,最后射到墙上,将 'x' 变成 '&' 3 模拟:仔细读题,搞清楚要做什么,就是i,j的移动,直到撞到 ...
组合数学(全排列)+DFS CSU 1563 Lexicography
题目传送门 1 /* 2 题意:求第K个全排列 3 组合数学:首先,使用next_permutation 函数会超时,思路应该转变, 4 摘抄网上的解法如下: 5 假设第一位是a,不论a是什么数,ax ...
js进阶 11-16 jquery如何查找元素的父亲、祖先和子代、后代
js进阶 11-16 jquery如何查找元素的父亲.祖先和子代.后代一.总结一句话总结:过滤或者查找的方法里面可以带参数进行进一步的选择. 1.parent()和parents()方法的区别是什 ...
PAT甲级1116 Come on! Let‘s C：[C++题解]哈希表、素数
文章目录题目分析题目来源题目分析来源:acwing 分析:使用哈希表存一下每个id获得奖,然后查询输出即可. ac代码 #include<bits/stdc++.h> using ...
CSU 1337 搞笑版费马大定理（2013湖南省程序设计竞赛J题）
题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1337 解题报告:虽然x和y的范围都是10^8,但是如果a 是大于1000的话,那么a^3 ...
LeetCode 多线程 1116. 打印零与奇偶数
1116. 打印零与奇偶数 Ideas 有几个线程就用几个信号量,最先开始的信号量初始化为1,其它初始化为0,然后根据条件判断实现同步. 多线程的问题好多都是:锁自己,解锁别人. Code from ...

CSU 1116 Kingdoms

CSU 1116 Kingdoms相关推荐

最新文章

热门文章