POJ 1422 && ZOJ 1525 Air Raid（最小路径覆盖）

Air Raid

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8127 Accepted: 4865

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

Dhaka 2002

题目大意：

    在一个有向无环图上，选择一些起始顶点，使得以这些顶点为起点可以到达所有点，求最少的起始顶点数。

解题思路：

    第一次遇到最小路径覆盖问题，其实这题也就是一个裸的最小路径覆盖。

    首先，我们把每个顶点拆成一个入点，一个出点。然后对于每条有向边，我们连接对应的入点和出点。这样我们就得到一张二分图。答案非常神奇的是初始顶点数-最大匹配值。

    当然这样写是有原理的：首先我们考虑最笨的方法，选择每一个顶点，这样满足题意（当然不一定是最少），接下来我们考虑怎么减少选择的顶点数。显然如果对于两个顶点一个有前驱，一个有后继，我们就可以由一个点走到另一个点，就可以少一个初始顶点。那么我们的任务就变成根据已有的边选择尽可能多的这样的点对，这就是我们熟悉的二分图最大匹配。所以答案就是顶点数-最大匹配值（全部顶点-能够去掉的最大值）。

AC代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <vector>#include <queue>using namespace std;#define INF 0x3f3f3f3f#define mem(a,b) memset((a),(b),sizeof(a))const int MAXN=120+3;const int MAXV=MAXN*2;int num_x;//左边顶点数vector<int> G[MAXV];//图的邻接表形式（左边顶点在前面，从0开始）int match_x[MAXV],match_y[MAXV];//顶点匹配对象int dis,dis_x[MAXV],dis_y[MAXV];//距离（用于多路增广）bool used[MAXV];bool searchP()//标记距离（用于多路增广）{    queue<int> que;    dis=INF;    mem(dis_x,-1);    mem(dis_y,-1);    for(int i=0;i<num_x;++i)        if(match_x[i]==-1)        {            que.push(i);            dis_x[i]=0;        }    while(!que.empty())    {        int u=que.front(); que.pop();        if(dis_x[u]>dis)            break;        for(int i=0;i<G[u].size();++i)        {            int v=G[u][i];            if(dis_y[v]==-1)            {                dis_y[v]=dis_x[u]+1;                if(match_y[v]==-1)                    dis=dis_y[v];                else                {                    dis_x[match_y[v]]=dis_y[v]+1;                    que.push(match_y[v]);                }            }        }    }    return dis!=INF;}bool dfs(int u)//dfs增广{    for(int i=0;i<G[u].size();++i)    {        int v=G[u][i];        if(!used[v]&&dis_y[v]==dis_x[u]+1)        {            used[v]=true;            if(match_y[v]!=-1&&dis_y[v]==dis)                continue;            if(match_y[v]==-1||dfs(match_y[v]))            {                match_y[v]=u;                match_x[u]=v;                return true;            }        }    }    return false;}int hopcroft_carp(){    int res=0;    mem(match_x,-1);    mem(match_y,-1);    while(searchP())    {        mem(used,0);        for(int i=0;i<num_x;++i)            if(match_x[i]==-1&&dfs(i))                ++res;    }    return res;}int N,M;int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&N,&M);        for(int i=0;i<N*2;++i)            G[i].clear();        num_x=N;        for(int i=0;i<M;++i)        {            int u,v;            scanf("%d%d",&u,&v);            G[u-1].push_back(v-1+N);        }        printf("%d\n",N-hopcroft_carp());    }        return 0;}