UVA 10859 有向无环图的动态规划

http://vjudge.net/vjudge/contest/view.action?cid=53516#problem/E

Description

Input: Standard In Output: Standard Out Next Generation Contest 1  Time Limit: 2 seconds Problem D Placing Lampposts

As a part of the mission �Beautification of Dhaka City�, the government has decided to replace all the old lampposts with new expensive ones. Since the new ones are quite expensive and the budget is not up to the requirement, the government has decided to buy the minimum number of lampposts required to light the whole city.

Dhaka city can be modeled as an undirected graph with no cycles, multi-edges or loops. There are several roads and junctions. A lamppost can only be placed on junctions. These lampposts can emit light in all the directions, and that means a lamppost that is placed in a junction will light all the roads leading away from it.

The �Dhaka City Corporation� has given you the road map of Dhaka city. You are hired to find the minimum number of lampposts that will be required to light the whole city. These lampposts can then be placed on the required junctions to provide the service. There could be many combinations of placing these lampposts that will cover all the roads. In that case, you have to place them in such a way that the number of roads receiving light from two lampposts is maximized.

Input

There will be several cases in the input file. The first line of input will contain an integer T(T<=30) that will determine the number of test cases. Each case will start with two integers N(N<=1000) and M( M<N) that will indicate the number of junctions and roads respectively. The junctions are numbered from 0 to N-1. Each of the next M lines will contain two integers a and b, which implies there is a road from junction a to b,
( 0<= a,b < N ) and a != b. There is a blank line separating two consecutive input sets.

Output

For each line of input, there will be one line of output. Each output line will contain 3 integers, with one space separating two consecutive numbers. The first of these integers will indicate the minimum number of lampposts required to light the whole city. The second integer will be the number of roads that are receiving lights from two lampposts and the third integer will be the number of roads that are receiving light from only one lamppost.

Sample Input

2 4 3 0 1 1 2 2 3 5 4 0 1 0 2 0 3 0 4

Sample Output

2 1 2 1 0 4

题目大意： （大白书p70）

           给你一个n个点m条边的无向无环图，在尽量少的节点上放灯，使得所有的边都被照亮，每盏灯将照亮以他为一个端点的所有边。在等的总数量最小的前提下，被两盏灯同时照亮的边数音尽量大。

解题思路：

        见大白书p71。（太多解释，我就不照着敲了）

#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;

vector<int>adj[1010];
int vis[1010][2],d[1010][2],n,m;

int dp(int i,int j,int f)//f是i的父亲节点
{
    if(vis[i][j])
        return d[i][j];
    vis[i][j]=1;
    int& ans=d[i][j];//引用的意思就是d[i][j]和ans完全等价，改变一个就相当于改变另一个
    //放灯总是合法的决策
    ans=2000;//灯的数量加1，x+2000
    for(int k=0; k<adj[i].size(); k++)
        if(adj[i][k]!=f)//这个判断非常重要！除了父节点之外的相邻节点才是子节点
            ans+=dp(adj[i][k],1,i);//注意这些节点的父节点是i;
    if(!j&&f>=0)ans++;//如果i不是根，且父节点没放灯，则x+1

    if(j||f<0)//i是根或者其父节点已放灯，i才可以不放灯
    {
        int sum=0;
        for(int k=0; k<adj[i].size(); k++)
            if(adj[i][k]!=f)
                sum+=dp(adj[i][k],0,i);
        if(f>=0)sum++;//如果i不是根，则x+1
        ans=min(ans,sum);
    }
    return ans;
}

int main()
{
    int T,a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0; i<n; i++) //adj里面保存着上一组数据的值，因此清空是必要的
            adj[i].clear();
        for(int i=0; i<m; i++)
        {
            scanf("%d%d",&a,&b);
            adj[a].push_back(b);
            adj[b].push_back(a);//无向图双向建边
        }
        memset(vis,0,sizeof(vis));
        int ans=0;
        for(int i=0; i<n; i++)
        {
            if(!vis[i][0])//新的一棵树
                ans+=dp(i,0,-1);//i是树根，因此父亲结点不存在为-1
        }
        printf("%d %d %d\n",ans/2000,m-ans%2000,ans%2000);//从x计算3个整数
    }
    return 0;
}