POJ1422 Air Raid 和 CH6902 Vani和Cl2捉迷藏

Air Raid

Language:Default

Air Raid

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9547		Accepted: 5696

Description

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.

Input

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.

Output

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.

Sample Input

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

Sample Output

2
1

Source

Dhaka 2002

有n个点和m条有向边，现在要在点上放一些伞兵，然后伞兵沿着图走，直到不能走为止
每条边只能是一个伞兵走过，问最少放多少个伞兵

题解

这就是最小路径点覆盖模板题。

拆点成出入点，连边，答案为n-最大匹配。

#include<iostream>
#include<bitset>
#include<cstring>
#include<vector>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
    for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
using namespace std;

co int N=301;
int n,m,f[N];
vector<int> e[N];
bitset<N> v;
bool dfs(int x){
	for(unsigned i=0;i<e[x].size();++i){
		int y=e[x][i];
		if(v[y]) continue;
		v[y]=1;
		if(!f[y]||dfs(f[y])){
			f[y]=x;
			return 1;
		}
	}
	return 0;
}
void Air_Raid(){
	read(n),read(m);
	for(int i=1;i<=n;++i) e[i].clear();
	for(int x,y;m--;){
		read(x),read(y);
		e[x].push_back(y);
	}
	memset(f,0,sizeof f);
	int ans=0;
	for(int i=1;i<=n;++i){
		v<<=n;
		ans+=dfs(i);
	}
	printf("%d
",n-ans);
}
int main(){
	for(int t=read<int>();t--;) Air_Raid();
	return 0;
}

6902 Vani和Cl2捉迷藏 0x60「图论」例题

描述

Vani和cl2在一片树林里捉迷藏。这片树林里有N座房子，M条有向道路，组成了一张有向无环图。N≤200，M≤30000。
树林里的树非常茂密，足以遮挡视线，但是沿着道路望去，却是视野开阔。如果从房子A沿着路走下去能够到达B，那么在A和B里的人是能够相互望见的。
现在cl2要在这N座房子里选择K座作为藏身点，同时Vani也专挑cl2作为藏身点的房子进去寻找，为了避免被Vani看见，cl2要求这K个藏身点的任意两个之间都没有路径相连。
为了让Vani更难找到自己，cl2想知道最多能选出多少个藏身点。

输入格式

输入数据的第一行是两个整数N和M。接下来M行，每行两个整数 x,y，表示一条从 x 到 y 的有向道路。

输出格式

输出一个整数K，表示最多能选取的藏身点个数。

在第二行输出 K 个空格分开的整数，表示选择的藏身点编号。如果有多个方案，输出任意一个即可。编号的输出顺序任意。

样例输入

7 5
1 2
3 2
2 4
4 5
4 6

样例输出

3
1 3 7

数据范围与约定

• 对于20% 的数据，N≤10，M<=20。
  对于60% 的数据, N≤100，M<=1000。
  对于100% 的数据，N≤200，M<=30000，1<=x,y<=N。

本题校验器(SPJ)

View

PrintAbout

01	#include <iostream>

02	#include <cstdio>

03	#include <cstring>

04	#include <algorithm>

05	#include <vector>

06	using namespace std;

07

//一些定义

08	const int ACCEPT = 0;

09	const int WRONG_ANSWER = 1;

10	//fstd 标准输出 fout 选手输出 fin 标准输入

11	FILE *fstd,*fout,*fin;

12	int n, m, ans, val;

13	bool v[210], f[210];

14	vector<int> ver[210];

15

16	bool dfs(int x) {

17

f[x] = 1;

18	if (v[x]) return true;

19	for (int i = 0; i < ver[x].size(); i++) {

20	int y = ver[x][i];

21	if (f[y]) continue;

22	if (dfs(y)) return true;

23

}

24	return false;

25

}

26

27

//执行比较操作。

28	bool DoCompare(){

29	fscanf(fin, "%d%d", &n, &m);

30	for (int i = 1; i <= m; i++) {

31

int x, y;

32	fscanf(fin, "%d%d", &x, &y);

33	ver[x].push_back(y);

34

}

35	fscanf(fstd, "%d", &ans);

36	fscanf(fout, "%d", &val);

37

// 答案不对

38	if (val != ans) return false;

39	for (int i = 1; i <= ans; i++) {

40	int x; fscanf(fout, "%d", &x);

41	// 输出的藏身点不合法或有重复

42	if (x < 1 || x > n || v[x]) return false;

43

v[x] = 1;

44

}

45	for (int i = 1; i <= n; i++) {

46	if (!v[i]) continue;

47	memset(f, 0, sizeof(f));

48

v[i] = 0;

49

// 能看到别的点

50	if (dfs(i)) return false;

51

v[i] = 1;

52

}

53	return true;

54

}

55

56	int main(int argc, char* argv[])

57

{

58	if(argc!=4){

59	printf("参数不足 %d",argc);

60	return -1;

61

}

62

63

//打开文件

64	if(NULL==(fstd=fopen(argv[1],"r"))){

65	return -1;

66

}

67	if(NULL==(fout=fopen(argv[2],"r"))){

68	return -1;

69

}

70	if(NULL==(fin=fopen(argv[3],"r"))){

71	return -1;

72

}

73

74	if(DoCompare()){

75	return ACCEPT;

76

}else{

77	return WRONG_ANSWER;

78

}

79

}

        </article>

题解

藏身点个数等于最小路径可重复点覆盖包含的路径条数。只需传递闭包，拆点跑二分图最大匹配，用点数减去它就行了。

证明见《进阶》，是一个利用了反证法的构造。

#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
    rg T data=0,w=1;rg char ch=getchar();
    for(;!isdigit(ch);ch=getchar())if(ch=='-') w=-w;
    for(;isdigit(ch);ch=getchar()) data=data*10+ch-'0';
    return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;

co int N=201;
bool cl[N][N];
int match[N],n,m;
bool vis[N],succ[N];
int hide[N];

bool dfs(int x){
	for(int i=1;i<=n;++i)
		if(cl[x][i]&&!vis[i]){
			vis[i]=1;
			if(!match[i]||dfs(match[i])){
				match[i]=x;
				return 1;
			}
		}
	return 0;
}
int main(){
	read(n),read(m);
	while(m--) cl[read<int>()][read<int>()]=1;
	for(int i=1;i<=n;++i) cl[i][i]=1;
	for(int k=1;k<=n;++k)
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				cl[i][j]|=cl[i][k]&cl[k][j];
	for(int i=1;i<=n;++i) cl[i][i]=0;
	// Maximum Matching on Split Bipartite Graph
	int ans=n;
	for(int i=1;i<=n;++i){
		memset(vis,0,sizeof vis);
		ans-=dfs(i);
	}
	printf("%d
",ans);
	for(int i=1;i<=n;++i) succ[match[i]]=1;
	for(int i=1,k=0;i<=n;++i)
		if(!succ[i]) hide[++k]=i;
	memset(vis,0,sizeof vis);
	for(bool modify=1;modify;){
		modify=0;
		for(int i=1;i<=ans;++i)
			for(int j=1;j<=n;++j)
				if(cl[hide[i]][j]) vis[j]=1;
		for(int i=1;i<=ans;++i)
			if(vis[hide[i]]){
				modify=1;
				while(vis[hide[i]]) hide[i]=match[hide[i]];
			}
	}
	for(int i=1;i<=ans;++i) printf("%d ",hide[i]);
	return 0;
}