hdu1151 二分图（无回路有向图）的最小路径覆盖 Air Raid
Air Raid
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3841 Accepted Submission(s): 2536
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.
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
2
4
3
3 4
1 3
2 3
3
3
1 3
1 2
2 3
Sample Output
2
1
Source
Recommend
题目大意：在一个城镇，有m个路口，和n条路，这些路都是单向的，而且路不会形成环，现在要弄一些伞兵去巡查这个城镇，伞兵只能沿着路的方向走，问最少需要多少伞兵才能把所有的路口搜一遍。
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; const int maxn = 125; bool vis[maxn]; //查询右集合中的点有没有被访问过 int link[maxn]; //link[i]表示右集合中的i点是由左集合中的哪个点连接的 int G[maxn][maxn]; //邻接矩阵 int x_cnt; int y_cnt; //左右集合的点的个数 bool find(int u) //用来寻找增广路 { for(int i = 1; i <= y_cnt; i++) //遍历右集合中的每个点 { if(!vis[i] && G[u][i]) //没有被访问过并且和u点有边相连 { vis[i] = true; //标记该点 if(link[i] == -1 || find(link[i])) { //该点是增广路的末端或者是通过这个点可以找到一条增广路 link[i] = u;//更新增广路 奇偶倒置 return true;//表示找到一条增广路 } } } return false;//如果查找了右集合里的所有点还没找到通过该点出发的增广路，该点变不存在增广路 } int solve() { int num = 0; memset(link, -1, sizeof(link));//初始化为-1表示 不与左集合中的任何元素有link for(int i = 1; i <= x_cnt; i++) //遍历左集合 { memset(vis, false, sizeof(vis));//每一次都需要清除标记 if(find(i)) num++;//找到一条增广路便num++ } return num; } int main() { int t1; scanf("%d",&t1); while(t1--){ scanf("%d",&x_cnt); y_cnt=x_cnt; memset(G,0,sizeof(G)); int x,y; int n; scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d%d",&x,&y); G[x][y]=1; } printf("%d\n",x_cnt-solve()); } return 0; }
