集训队专题（6）1001 Air Raid

Air Raid

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4378    Accepted Submission(s): 2920

Problem 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

2433 41 32 3331 31 22 3

 

Sample Output

21

 

Source

Asia 2002, Dhaka (Bengal)

 

题意：一个城镇中有n个十字路口和m条单项的路径，图是无环图，现在要派一些伞兵去这些成寿寺，要到达所有的路口，每个在一个路口着陆了的伞兵可以沿着街去到其他路口；我们的任务就是求出去执行任务的伞兵最少可以使几个；

题目相当于要我们找出需要最少的路径，使得我们的路径包含了所有的顶点，也就是要我们求出二分图问题中的最小路径覆盖，根据二分图的定理，最小路径覆盖 = 图的顶点数 - 最大匹配数，问题又回到了让我们求出最大匹配数上，匈牙利算法的模板一套过来，就AC啦~

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100+10;int uN,vN;int g[maxn][maxn],linker[maxn];bool used[maxn];bool dfs(int u){int v;for(v=1; v<=vN; v++)if(g[u][v] && !used[v]){used[v] = true;if(linker[v]==-1 || dfs(linker[v])){linker[v] = u;return true;}}return false;}int hungary(){int res = 0;int u;memset(linker,-1,sizeof(linker));for(u=1; u<=uN; u++){memset(used,0,sizeof(used));if(dfs(u)) res++;}return res;}int main(){int m,t,ans;int u,v;scanf("%d",&t);while(t--){scanf("%d%d",&uN,&m);vN = uN;memset(g,0,sizeof(g));for(int i=0; i<m; i++){scanf("%d%d",&u,&v);g[u][v] = 1;}ans = uN-hungary();printf("%d\n",ans);}return 0;}