poj 3659 Cell Phone Network——树形dp最��点覆盖所有点 �Ҏ��配集

Mon, 11 Apr 2011 14:06:00 GMT

Cell Phone Network

Description

Farmer John has decided to give each of his cows a cell phone in hopes to encourage their social interaction. This, however, requires him to set up cell phone towers on his N (1 ≤ N ≤ 10,000) pastures (conveniently numbered 1..N) so they can all communicate.

Exactly N-1 pairs of pastures are adjacent, and for any two pastures A and B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B) there is a sequence of adjacent pastures such that A is the first pasture in the sequence and B is the last. Farmer John can only place cell phone towers in the pastures, and each tower has enough range to provide service to the pasture it is on and all pastures adjacent to the pasture with the cell tower.

Help him determine the minimum number of towers he must install to provide cell phone service to each pasture.

Input

* Line 1: A single integer: N
* Lines 2..N: Each line specifies a pair of adjacent pastures with two space-separated integers: A and B

Output

* Line 1: A single integer indicating the minimum number of towers to install

Sample Input

Sample Output

题意�Q�农民的n个地斚w��要徏电话亭，A有电话，则A盔R��的点可以要电话亭也可以不需要电话亭�?/span>
分析�Q�最��点覆盖�?/span>
分三�U�情况：
dp[u][0]u��Z��?/span>
dp[u][1]u不徏亭，不依赖父节点�Q�但u的孩子必��L��一个有�?/span>
dp[u][2]u不徏亭，依赖于父节点�Q�只要父节点能徏个亭�Q��?/span>
代码�Q?/span>

#include <stdio.h>
#include <stdlib.h>
#define inf (1 << 29)
#define Min(a, b) (a) < (b) ? (a) : (b)
#define maxn 21000
struct T
{
    int v, next;
}fn[maxn];
int th;
int g[maxn], dp[maxn][3], degree[maxn];
void add(int u, int v)
{
    fn[th].v = v, fn[th].next = g[u], g[u] = th++;
}
void dfs(int u, int f)
{
    if (u != 1 && degree[u] == 1)
    {
        dp[u][0] = 1;
        dp[u][1] = inf;
        dp[u][2] = 0;
        return;
    }
    dp[u][0] = 1;
    int i, v, sign, min, del;
    for (i = g[u], sign = 0, min = inf; i != -1; i = fn[i].next)
    {
         v = fn[i].v;
        if (v == f)
        {
            continue;
        }
        dfs(v, u);
        //如果取u�Q�则孩子v取MIN(的可取[v被自��p��盖]�Q�依赖孩子的孩子[表示v不用已经被覆盖，不依赖u]�Q�依赖父节点u[因�ؓ父节点现在取了])
        dp[u][0] += Min(Min(dp[v][0], dp[v][1]), dp[v][2]);
        if (dp[v][0] <= dp[v][1])//u不取也不依赖于父节点�Q�则u依赖于孩子，保证一个孩子被取�?/span>
        {
            dp[u][1] += dp[v][0];
            sign = 1;//选了一个有电话的，标志1
        }
        else//选了一个孩子没电话的，要保存min�Q�如果sign没标志过�Q�就得利用有电话的最��值得那个人�?/span>
        {
            dp[u][1] += dp[v][1];
            if (min > dp[v][0])
            {
                min = dp[v][0];
                del = dp[v][1];
            }
        }
        dp[u][2] += Min(dp[v][0], dp[v][1]);//u不取�Q�则u依赖于父节点�Q�则取v不依赖与u的最优情�?/span>
    }
    if (!sign)
    {
        dp[u][1] += min - del;
    }
}
int main()
{
    int n, u, v, i, j;
    while (scanf("%d", &n) - EOF)
    {
        th = 0;
        for (i = 0; i <= n; i++)
        {
            degree[i] = 0;
            g[i] = -1;
            dp[i][0] = dp[i][1] = dp[i][2] = 0;
        }
        for (i = 1; i < n; i++)
        {
            scanf("%d%d", &u, &v);
            add(u, v);
            add(v, u);
            degree[u]++;
            degree[v]++;
        }
        if (n == 1)
        {
            printf("0\n");
            continue;
        }
        dfs(1, 0);
        printf("%d\n", Min(dp[1][0], dp[1][1]));
    }
    return 0;
}

火碳�?/a> 2011-04-11 22:06 发表评论

Tue, 27 Apr 2010 15:05:00 GMT

Strategic Game

Time Limit: 10 Seconds Memory Limit: 32768 KB

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob has to put for a given tree.

The input file contains several data sets in text format. Each data set represents a tree with the following description:

the number of nodes
the description of each node in the following format
node_identifier:(number_of_roads) node_identifier1 node_identifier2 ... node_identifier
or
node_identifier:(0)

The node identifiers are integer numbers between 0 and n-1, for n nodes (0 < n <= 1500). Every edge appears only once in the input data.

For example for the tree:

the solution is one soldier ( at the node 1).

The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following table:

Input

4
0:(1) 1
1:(2) 2 3
2:(0)
3:(0)
5
3:(3) 1 4 2
1:(1) 0
2:(0)
0:(0)
4:(0)

Output

1
2

题意�Q�给出树的结点之间的关系,�?:(2) 2 3 �Q�表�C?�?的父亲都�?.�?点设点，则其��L��?�?都可被覆盖，求出最��要多少个点才能覆盖整棵树。即最��顶点覆盖�?/p>

分析�Q�树的各点之间只有一条边�Q�而且树明昄��Ҏ��是有层�ơ的数据�l�构。f[u][0]表示以u为根且u不加入覆盖集的最��顶点覆盖，f[u][1]表示以u为根且u加入覆盖集的最��顶点覆盖。每个点都有两个状态，��是加入与不加入覆盖集�?/p>

最优子�l�构�Q�要知道f[u][0]=只需知道u的子节点个数�Q�要知道f[u][1]�Q�则要知道分别以u的孩子v们�ؓ根的最��顶点覆盖，对这�U�情况，每个孩子可加入也可不加入覆盖集（取最优的�Q�，以�ؓ树有层次�Q�所以要知道栚w��Q�就要知道孩子的最��顶点覆盖，孩子的最��顶点覆盖跟父亲没有关系�?/p>

子问题重叠：�q�里没有子问题重叠�?/p>

边界问题�Q�每个节点对应两�U�边界问题，卛_��处理一个节�Ҏ��Q�有取和不取两种情况�?/p>

子问题独立：对于u的孩子v1�Q�v2�Q�v1的孩子跟v2的孩子毫无关�p�R��这�U�性质延��到u的所有孩子上�Q�即每个孩子的解互相没有关系�?/p>

代码�Q?/p>

#include<stdio.h>
#define Min(a, b) a < b ? a : b
#define maxn 1510
int up[maxn], f[maxn][2], n;
void dp(int u)
{
    if (f[u][1] != -1)
    {
        return;
    }
    int i, v;
    for (v = 0, f[u][1] = 1, f[u][0] = 0; v < n; v++)//对于每个点，递归到每个孩子解决问题后才能解决自��n问题
    {
        if (up[v] == u)
        {
            dp(v);
            f[u][1] += Min(f[v][0], f[v][1]);
            f[u][0] += f[v][1];
        }
    }
}
int main()
{
    int i, u, v, m, root;
    while (scanf("%d", &n) != EOF)
    {
        for (i = 0; i < n; i++)
        {
            up[i] = -1;
            f[i][0] = f[i][1] = -1;
        }
        for (i = 0; i < n; i++)
        {
            scanf("%d:(%d)", &u, &m);
            while (m--)
            {
                scanf("%d", &v);
                up[v] =  u;
            }
        }
        root = 0;
        while (up[root] != -1)
        {//查找根结�?nbsp;
            root = up[root];
        }
        dp(root);
        printf("%d\n", Min(f[root][1], f[root][0]));
    }
    return 0;
}

火碳�?/a> 2010-04-27 23:05 发表评论

国产一区二区精品久久,亚洲国产视频一区,国产亚洲一级

poj 3659 Cell Phone Network——树形dp最���点覆盖所有点 �Ҏ��配集

poj 3659 Cell Phone Network——树形dp最��点覆盖所有点 �Ҏ��配集