2013
12-03

# Air Raid

Consider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting from an intersection and walking through town’s streets you can never reach the same intersection i.e. the town’s streets form no cycles.

With these assumptions your task is to write a program that finds the minimum number of paratroopers that can descend on the town and visit all the intersections of this town in such a way that more than one paratrooper visits no intersection. Each paratrooper lands at an intersection and can visit other intersections following the town streets. There are no restrictions about the starting intersection for each paratrooper.

Your program should read sets of data. The first line of the input file contains the number of the data sets. Each data set specifies the structure of a town and has the format:

no_of_intersections
no_of_streets
S1 E1
S2 E2
……
Sno_of_streets Eno_of_streets

The first line of each data set contains a positive integer no_of_intersections (greater than 0 and less or equal to 120), which is the number of intersections in the town. The second line contains a positive integer no_of_streets, which is the number of streets in the town. The next no_of_streets lines, one for each street in the town, are randomly ordered and represent the town’s streets. The line corresponding to street k (k <= no_of_streets) consists of two positive integers, separated by one blank: Sk (1 <= Sk <= no_of_intersections) – the number of the intersection that is the start of the street, and Ek (1 <= Ek <= no_of_intersections) – the number of the intersection that is the end of the street. Intersections are represented by integers from 1 to no_of_intersections.

There are no blank lines between consecutive sets of data. Input data are correct.

The result of the program is on standard output. For each input data set the program prints on a single line, starting from the beginning of the line, one integer: the minimum number of paratroopers required to visit all the intersections in the town.

2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3

2
1

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 110;
int n,m;
int g[maxn][maxn];
int match[maxn];
bool vis[maxn];
bool dfs(int cur){

for(int i=1;i<=n;i++){
if(g[cur][i]&&!vis[i]){
vis[i]=true;
int t=match[i];

if(t==-1||dfs(t)){
match[i]=cur;
return true;
}

}
}
return false;

}
int hungary(){
for(int i=1;i<=n;i++)  match[i]=-1;
int res=0;
for(int i=1;i<=n;i++){
memset(vis,false,sizeof(vis));
if(dfs(i))
res++;
}
return res;
}
int main(){
int t,x,y;
scanf("%d",&t);
while(t--){
memset(g,0,sizeof(g));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++){
scanf("%d%d",&x,&y);
g[x][y]=1;
}
printf("%d\n",n-hungary());
}
return 0;

}

1. #include <stdio.h>
int main(void)
{
int arr[] = {10,20,30,40,50,60};
int *p=arr;
printf("%d,%d,",*p++,*++p);
printf("%d,%d,%d",*p,*p++,*++p);
return 0;
}

为什么是 20,20,50,40,50. 我觉得的应该是 20,20,40,40,50 . 谁能解释下？

2. 思路二可以用一个长度为k的队列来实现，入队后判断下队尾元素的next指针是否为空，若为空，则出队指针即为所求。

3. 可以根据二叉排序树的定义进行严格的排序树创建和后序遍历操作。如果形成的排序树相同，其树的前、中、后序遍历是相同的，但在此处不能使用中序遍历，因为，中序遍历的结果就是排序的结果。经在九度测试，运行时间90ms，比楼主的要快。