I would let the 18-year third title
do not really know the shortest path and minimum spanning tree, then you can see more of this
https://blog.csdn.net/spaceyqy/article/details/39024675
Watermelon watermelon Uncle Wang last year a lot of money. See decent income, this year he again opened up the n watermelon.
In order to give a smooth doused his n watermelon to water, for each of the watermelon he may choose in the local wells, repair the pipeline can also be from another melon (melon may hit this well; it may not play well, he also drew water from other Guardia) lead over the water.
Of course, the cost of drilling and pipeline repair a difference. Wells are known in the i-th element wi watermelon I take, between the first i, j watermelon takes to repair the pipeline pi, j element.
The question now is: Uncle Wang in order to make all Guardia are doused with water, at least need to spend much money (the cost of drilling wells and pipelines and repair)?
Because Guardia more, Uncle Wang can not be selected in which (a) Guardia wells, repair the pipeline between what the watermelon.
Please help Uncle Wang programming decisions, the value of minimum cost.
（1<=N<=300，1<=wi<=100000；pi,i=0，1<=pi,j=pj,i<=100000）
input
Line 1, a positive integer n, the number of representative watermelon.
The following n rows sequentially given integers w1 ... wn (watermelon drilling costs of each).
Followed by a matrix of n * n is an integer, the i-th row of the matrix represents the number PI, j of the j-th column (the cost of establishing the pipe between two watermelon). There are between two numbers each line separated by a space.
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
//瓜农王大爷去年种西瓜赚了不少钱。看到收入不错，今年他又重新开辟了n个西瓜地。
//为了能给他的n个西瓜地顺利的浇上水，对于每个西瓜地他可以选择在本地打井，也可以修管道从另一个瓜地（这个瓜地可能打了井；也可能没打井，他的水也是从其他瓜地引来的）将水引过来。
//当然打井和修管道的费用有差别。已知在第i个西瓜地打井需要耗费wi元，在第i、j个西瓜地之间修管道需要耗费pi,j元。
//现在的问题是：王大爷要想使所有瓜地都被浇上水，至少需要花费多少钱（打井与修管道的费用和）？
//由于瓜地较多，王大爷无法选择在哪些（个）瓜地打井，哪些西瓜地之间修管道。
//请你编程帮王大爷做出决策，求出最小费用。
//这道题和我之前写过的类似，就是求最短路径，可以将打井的钱，当做是从原点到该水管的距离，这样就增加了一些距离
//可以创建一个数组来存储第一位是起点，第二位是终点，值是对应距离
//用到并查集
int a[100];//用来存对应点的祖宗
int zuzong(int x){
//找到某个点的祖宗
if(a[x]==x)
return x;
else
return zuzong(a[x]);
}
bool check(int x,int y){
//判断两点是否同一个祖宗，也就是在不在同一个集合
return zuzong(x)==zuzong(y);
}
void merge_1(int x,int y){
//合并两个点到同一个祖宗，前提是他们不是同一个祖宗
a[zuzong(x)]=zuzong(y);
}
int dis[100][100];
int i,j;
int n;//西瓜地的数量
struct Y{
int s,e,dis;//对应起点，终点，和对应距离
}Gua[100];
bool cmp(Y a,Y b){
return a.dis<b.dis;//距离从小到大排序
}
int main(){
scanf("%d",&n);
int temp[100][100];//中间数组
for(i=1;i<=n;i++){
Gua[i].s=0;
Gua[i].e=i;
scanf("%d",&Gua[i].dis);
}
int m=n+1;//因为前面n个点已经记录了好了距离，所以需要从第n+1个点开始
for(i=1;i<=n;i++)
for(j=1;j<=n;j++){
scanf("%d",&temp[i][j]);
if(i<j){
Gua[m].s=i;
Gua[m].e=j;
Gua[m].dis=temp[i][j];//修管道的费用,只输出单方向的，让无向图变成有向的
m++;
}
}
//对数组进行花费从小到大排序
sort(Gua+1,Gua+m+1,cmp);//n是西瓜地数量，c是边数,默认从小到大
//再将每个点，默认祖宗初始设置为自己
for(i=0;i<=n;i++)
a[i]=i;
//开始合并
int sum=0;
for(i=1;i<=m;i++)//一共m条边
if(!check(Gua[i].s,Gua[i].e)){//没有合并，那就合并，并且记录对应花费
sum+=Gua[i].dis;
merge_1(Gua[i].s,Gua[i].e);
}
printf("%d",sum);
}
first question
Problem Description
A group of people go to the cinema. But the cinema has a very strange requirement: adult seat number can only be assigned odd numbers, children under 18 can only be assigned the number is even seat number.
Input
There are n individual input to the movies (1 <= n <1000) , the next n individual input seat number, seat number for each space-separated
Output
The ratio of output in order to see the number of adult movies and adult share of all people, the proportion of the number of minors and minor share of all people, the calculated ratio of two decimal places, each output separated by a space
Sample Input
8 13 12 10 8 3 24 21 19
Sample Output
4 0.50 4 0.50
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
//第一题
//Problem Description
//有一群人去电影院看电影。但电影院有个很奇怪的规定：成人只能分到数字奇数座位号，未满18岁的儿童只能分到数字为偶数的座位号。
//Input
//输入共有n个人去看电影(1<=n<1000)，接下来输入n个人的座位号，每个座位号用空格隔开
//Output
//依次输出此次看电影成人的人数以及成人在所有人中所占的比例、未成年人的人数以及未成年人在所有人中所占的比例，计算出的比例保留两位小数，每个输出用空格隔开
//Sample Input
//8 13 12 10 8 3 24 21 19
//Sample Output
//4 0.50 4 0.50
int main(){
int i,n;//输入的人数
int sum1=0;
int sum2=0;
int a[100];
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d",&a[i]);
if((a[i]%2)==0)
sum2++;
else
sum1++;
}
printf("%d %.2f %d %.2f",sum1,(1.0*sum1)/n,sum2,(1.0*sum2)/n);
}
Problem Description
There is a large container, which is now added to the plurality of plumb wood, wood sit top of each labeled (i, yi), as shown in FIG
Input
Was added a total of n input boards (1 <= n <1000) , represented by n digital input next added to the height of the n boards, each separated by a space the height of
Output
The output of the largest number of containers can hold a volume of water (the inclination of the container does not allow)
the Sample the Input
8 1 8 6 4 5 3 7 2
Sample Output
35
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
//第二题，给n模板，输入n个值代表模板的高度，求最大容积
//我感觉就直接暴力，两次for循环，i从0-n-1 然后J 从i+1 到n-1 直接求最大值
int main(){
int i,j,n;
scanf("%d",&n);
int a[100];
for(i=0;i<n;i++)
scanf("%d",&a[i]);
int v,mmax=0;
for(i=0;i<n;i++)
for(j=i+1;j<n;j++){
v=(j-i)*min(a[i],a[j]);
if(v>mmax)
mmax=v;
}
printf("%d",mmax);
}
Problem Description
A group of people, to tell you now friendship between them. If A and B are friends, friends of B is C, then A and C are also friends. Friend relationships are bidirectional, that A and B are friends, A and B are also friends. Then A, B, C in the same "circle of friends"
the Input
First inputs n, m denotes n-personal, m relationship (1 <= n <2000) , followed m rows each row represents a relationship (each digit representing a person)
the Output
In the case where the current input output relationship between the number "circle of friends" in
the Sample the Input
. 8. 7
. 1 2
2. 4
. 4. 1
. 5. 7
. 4. 3
. 6 2
. 7. 8
the Sample the Output
2
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
//第四题有点像是连通图， 求连通图的个数。那就是用dfs或者bfs,或者是用并查集，然后用一个coun++
struct Y{
int s,e;
}fri[100];
int a[100];
int zuzong(int x){
if(x==a[x])
return x;
else
return zuzong(a[x]);
}
bool check(int x,int y){
return zuzong(x)==zuzong(y);
}
void merge_1(int x,int y){
a[zuzong(x)]=zuzong(y);
}
int main(){
int n,m;//n个人，m种关系
int i,c=1;//初始值为1，假设都连通
scanf("%d%d",&n,&m);
for(i=0;i<m;i++){
scanf("%d%d",&fri[i].s,&fri[i].e);
}
//先赋初始值
for(i=1;i<=n;i++)
a[i]=i;
for(i=0;i<m;i++){
if(!check(fri[i].s,fri[i].e)){//就是靠并查集来记录，有哪些是不一样的
merge_1(fri[i].s,fri[i].e);
}
}
//这个时候已经全都分好类了，然后就开始筛选吧，
//我想到一个办法就是对数组进行排序，然后找到不一样的值
//创建一个数组来存祖宗
int temp[100];
for(i=1;i<=n;i++)
temp[i]=zuzong(i);
//for(i=1;i<=n;i++)
//printf("%d ",temp[i]);
sort(temp+1,temp+1+n);
int flag=temp[1];
for(i=2;i<=n;i++){
if(temp[i]!=flag){
flag=temp[i];
c++;
}
}
printf("%d",c);
}
