HDU-1151 Air Raid(最小パスオーバーライド)

13010 ワード

Air Raid
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5826 Accepted Submission(s): 3919
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

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

Sample Output

2 1

Source

Asia 2002, Dhaka (Bengal)

Recommend

Ignatius.L

题意：求最小路径覆盖

解题思路：

有向无环图（DAG）的最小路径覆盖

DAG的最小路径覆盖

定义：在一个有向图中，找出最少的路径，使得这些路径经过了所有的点。

最小路径覆盖分为最小不相交路径覆盖和最小可相交路径覆盖。

最小不相交路径覆盖：每一条路径经过的顶点各不相同。如图，其最小路径覆盖数为3。即1->3>4，2，5。

最小可相交路径覆盖：每一条路径经过的顶点可以相同。如果其最小路径覆盖数为2。即1->3->4，2->3>5。

特别的，每个点自己也可以称为是路径覆盖，只不过路径的长度是0。

DAG的最小不相交路径覆盖

算法：把原图的每个点V拆成Vx和Vy两个点，如果有一条有向边A->B，那么就加边Ax−>By。这样就得到了一个二分图。那么最小路径覆盖=原图的结点数-新图的最大匹配数。

证明：一开始每个点都是独立的为一条路径，总共有n条不相交路径。我们每次在二分图里找一条匹配边就相当于把两条路径合成了一条路径，也就相当于路径数减少了1。所以找到了几条匹配边，路径数就减少了多少。所以有最小路径覆盖=原图的结点数-新图的最大匹配数。

因为路径之间不能有公共点，所以加的边之间也不能有公共点，这就是匹配的定义。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long int ll;
const int MAXN = 505;

vector G[MAXN];
int N, M;

//     
bool vis[MAXN];
int Y[MAXN];
bool hungry(int x)
{
    for (int i = 0; i < G[x].size(); i++)
    {
        if (vis[G[x][i]] == 0)
        {
            vis[G[x][i]] = 1;
            if (Y[G[x][i]] == -1 || hungry(Y[G[x][i]]))
            {
                Y[G[x][i]] = x;
                return true;
            }
        }
    }
    return false;
}

int main()
{
    int t;
    scanf("%d", &t);

    while (t--)
    {
        scanf("%d%d", &N, &M);
        for (int i = 1; i <= N; i++)
        {
            G[i].clear();
        }
        int a, b;
        for (int i = 0; i < M; i++)
        {
            scanf("%d%d", &a, &b);
            G[a].push_back(b);
        }

        int ans = 0;
        memset(Y, -1, sizeof(Y));
        for (int i = 1; i <= N; i++)
        {
            memset(vis, 0, sizeof(vis));
            if (hungry(i))
                ans++;
        }

        printf("%d
", N - ans);
    }

    return 0;
}

JanomeとMecabを使って形態素解析してみる

『thinking in Java』を読むのはすべて対象です