POJ 1422 Air Raid (二分匹配)
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Air Raid
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5690 Accepted: 3393
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.
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.
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
Dhaka 2002
题目要求的是最少降落多少个人,使得所有的intersections都能被vist。对于样例3->4,1->3,2->3,分别降落在1和2就可以了。额,顶着自己画的图看了半天,突然发现匈牙利算法每次都是找可增广路,那么这题可以看成每次找不到增广路....
#include <iostream>#include <cstdio>#include <vector>#include <cstring>using namespace std;const int maxn=250;vector<int>edge[maxn];bool vis[maxn];int inum,snum,match[maxn];bool find(int now){ int i,v; for(i=0; i<edge[now].size(); ++i) if(!vis[v=edge[now][i]]) { vis[v]=true; if(match[v]==-1||find(match[v])) { match[v]=now; return true; } } return false;}int main(){ int set,u,v; cin>>set; while(set--) { cin>>inum>>snum; memset(match,-1,sizeof(match)); for(int i=1; i<=inum; i++) edge[i].clear(); for(int i=1; i<=snum; i++) { cin>>u>>v; edge[u].push_back(v); } int ans=0; for(int i=1; i<=inum; i++) { memset(vis,0,sizeof(vis)); if(!find(i)) ans++; } cout<<ans<<endl; } return 0;}
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