POJ 1848 Tree

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3506		Accepted: 1204

Description

Consider a tree with N vertices, numbered from 1 to N. Add, if it is possible, a minimum number of edges thus every vertex belongs to exactly one cycle.

Input

The input has the following structure: 
N 
x(1) y(1) 
x(2) y(2) 
... 
x(N-1) y(n-1) 
N (3 <= N <=100) is the number of vertices. x(i) and y(i) (x(i), y(i) are integers, 1 <= x(i), y(i) <= N) represent the two vertices connected by the i-th edge. 

Output

The output will contain the value -1 if the problem doesn't have a solution, otherwise an integer, representing the number of added edges.

Sample Input

7
1 2
1 3
3 5
3 4
5 6
5 7

Sample Output

2

Source

Romania OI 2002

树形DP啊，可是知道也推不出公式来，。无奈的看了解题报告。发现前方的道路还非常漫长啊，这思路神了

http://www.cnblogs.com/vongang/archive/2012/08/12/2634763.html

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#define N 110
#define INF 150
using namespace std;
struct num
{
    int y,next;
}a[N*N];
int b[N],Top;
int dp[N][3];
bool ch[N];
int main()
{
    //freopen("data.txt","r",stdin);
    void addeage(int x,int y);
    void dfs(int x);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        Top = 0;
        memset(b,-1,sizeof(b));
        for(int i=1;i<=n-1;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            addeage(x,y);
            addeage(y,x);
        }
        for(int i=1;i<=n;i++)
        {
            dp[i][0] = dp[i][1] = dp[i][2] = INF;
        }
        memset(ch,false,sizeof(ch));
        dfs(1);
        //cout<<dp[3][1]<<endl;
        if(dp[1][0]==INF)
        {
            printf("%d\n",-1);
        }else
        {
            printf("%d\n",dp[1][0]);
        }
    }
    return 0;
}
void addeage(int x,int y)
{
    a[Top].y = y;
    a[Top].next = b[x];
    b[x] = Top++;
}
void dfs(int x)
{
    ch[x] = true;
    bool uv = true;
    for(int i=b[x];i!=-1;i=a[i].next)
    {
        int y = a[i].y;
        if(!ch[y])
        {
            uv = false;
            break;
        }
    }
    if(uv)
    {
        dp[x][1] =0;
        return ;
    }
    dp[x][1] = 0;
    int sum = 0;
    bool ch2[N];
    memset(ch2,false,sizeof(ch2));
    for(int i=b[x];i!=-1;i=a[i].next)
    {
        int y =  a[i].y;
        if(!ch[y])
        {
            dfs(y);
            ch2[y] = true;
            dp[x][1]+=dp[y][0];
            sum+=dp[y][0];
        }
    }
    if(dp[x][1]>=INF)
    {
        dp[x][1] = INF;
    }
    for(int i=b[x];i!=-1;i=a[i].next)
    {
        int y = a[i].y;
        if(!ch2[y])
        {
            continue;
        }
        int w = sum-dp[y][0]+min(dp[y][1],dp[y][2]);
        if(w>=INF)
        {
            w = INF;
        }
        dp[x][2] = min(dp[x][2],w);
        w = sum-dp[y][0] + dp[y][2]+1;
        if(w>=INF)
        {
            w = INF;
        }
        dp[x][0] = min(dp[x][0],w);
    }
    for(int i=b[x];i!=-1;i=a[i].next)
    {
        int y1 = a[i].y;
        if(!ch2[y1])
        {
            continue;
        }
        for(int j=b[x];j!=-1;j=a[j].next)
        {
            int y2 = a[j].y;
            if(y1==y2||!ch2[y2])
            {
                continue;
            }
            int w = sum-dp[y1][0] - dp[y2][0] + min(dp[y1][1],dp[y1][2]) + min(dp[y2][1],dp[y2][2])+1;
            if(w>=INF)
            {
                w = INF;
            }
            dp[x][0] = min(dp[x][0],w);
        }
    }
}