poj 3345 Bribing FIPA(树形DP&输入控制)
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Bribing FIPA
Description There is going to be a voting at FIPA (Fédération Internationale de Programmation Association) to determine the host of the next IPWC (International Programming World Cup). Benjamin Bennett, the delegation of Diamondland to FIPA, is trying to seek other delegation's support for a vote in favor of hosting IWPC in Diamondland. Ben is trying to buy the votes by diamond gifts. He has figured out the voting price of each and every country. However, he knows that there is no need to diamond-bribe every country, since there are small poor countries that take vote orders from their respected superpowers. So, if you bribe a country, you have gained the vote of any other country under its domination (both directly and via other countries domination). For example, if C is under domination of B, and B is under domination of A, one may get the vote of all three countries just by bribing A. Note that no country is under domination of more than one country, and the domination relationship makes no cycle. You are to help him, against a big diamond, by writing a program to find out the minimum number of diamonds needed such that at least m countries vote in favor of Diamondland. Since Diamondland is a candidate, it stands out of the voting process. Input The input consists of multiple test cases. Each test case starts with a line containing two integers n (1 ≤ n ≤ 200) and m (0 ≤ m ≤ n) which are the number of countries participating in the voting process, and the number of votes Diamondland needs. The next n lines, each describing one country, are of the following form: CountryName DiamondCount DCName1 DCName1 ... CountryName, the name of the country, is a string of at least one and at most 100 letters and DiamondCount is a positive integer which is the number of diamonds needed to get the vote of that country and all of the countries that their names come in the list DCName1 DCName1 ... which means they are under direct domination of that country. Note that it is possible that some countries do not have any other country under domination. The end of the input is marked by a single line containing a single # character. Output For each test case, write a single line containing a number showing the minimum number of diamonds needed to gain the vote of at least m countries. Sample Input 3 2 Aland 10 Boland 20 Aland Coland 15 # Sample Output 20 Source
Tehran 2006
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题意：
一人去竞选。共有n(1<=n<=200)个国家参与投票。通过贿赂可以得到他们的选票。而贿赂不同的国家所花费的代价也不同。国家之间有大哥和小弟的关系。如果贿赂了它的大哥就不用贿赂小弟了。因为小弟会跟着大哥投票。先问你
得到m个国家选票所花费的最小代价。
思路：
这题树形DP的模型还是很简单的。比较难弄的是输入。因为它没告诉你它后面有多少小弟。而且读n,m也必须用字符
数组读。因为它是用'#'来结束输入的。。--||。虽然花了不少功夫。但是也学了不少东西。还是值得的。
处理好输入后面就很简单了。可以用国家的名字和标号建立一个映射。然后用标号建单向边就行了。这次换了下风格。用vector存的边。树建好后就可以DP了。要注意的是题目给的是森林。所以增加0号节点把它变为有根树。
dp[i][j]表示。节点i获得j票的最小代价。
dp[i][j]=min(dp[i][j],dp[i][k]+dp[son][j-k])。son为i的儿子结点。
还有很多细节。
详细见代码：
#include <iostream>
#include<algorithm>
#include<sstream>//for stringstream
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<map>
using namespace std;
const int maxn=250;
const int INF=0x3f3f3f3f;
int n,m;
int cost[maxn];
int dp[maxn][maxn];
char buff[250110];
vector<int> edge[maxn];
map<string,int> mp;
string name;
bool nroot[maxn];// if the node is root nroot=0;
int dfs(int u)
{
int i,j,k,v,sz,lim,t,tot=1;
sz=edge[u].size();
dp[u][0]=0;
for(i=0;i<sz;i++)
{
v=edge[u][i];
t=dfs(v);
tot+=t;
for(j=tot;j>0;j--)
{
lim=min(j,t);
for(k=1;k<=lim;k++)
dp[u][j]=min(dp[u][j],dp[u][j-k]+dp[v][k]);
}
}
for(i=1;i<=tot;i++)//it's very important!
dp[u][i]=min(dp[u][i],cost[u]);
return tot;//the size of subtree.
}
int main()
{
int i,cnt,u,v;
stringstream s;//attention!
while(gets(buff))// gets() if empty return NULL.
{
if(buff[0]=='#')//a big track.
break;
sscanf(buff,"%d%d",&n,&m);
cnt=1;// 0 is the root
cost[0]=INF;
mp.clear();
memset(nroot,0,sizeof nroot);
memset(dp,0x3f,sizeof dp);
for(i=0;i<=n;i++)
edge[i].clear();
for(i=0;i<n;i++)
{
gets(buff);
s.clear();
s.str(buff);//a very useful method.
s>>name;
if(!mp[name])//build a reflect.
mp[name]=cnt++;
u=mp[name];
s>>cost[u];
while(s>>name)
{
if(!mp[name])
mp[name]=cnt++;
v=mp[name];
nroot[v]=true;
edge[u].push_back(v);//build edges.
}
}
for(i=1;i<cnt;i++)
if(!nroot[i])//if the node is root.
edge[0].push_back(i);
dfs(0);
printf("%d\n",dp[0][m]);
}
return 0;
}