問題描述

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.

輸入格式

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.

輸出格式

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.

樣例輸入

3 2
Aland 10
Boland 20 Aland
Coland 15

樣例輸出

20

題目大意

有一棵n個(gè)節(jié)點(diǎn)的樹，給一個(gè)節(jié)點(diǎn)染色的話，該節(jié)點(diǎn)的子樹的所有節(jié)點(diǎn)都會(huì)被染色。而給每一個(gè)節(jié)點(diǎn)染色都有一個(gè)代價(jià)，要求用最少的代價(jià)使被染色的節(jié)點(diǎn)數(shù)量不少于m。

解析

最直接的想法是設(shè)\(f[i][k]\)表示有i個(gè)節(jié)點(diǎn)被染色時(shí)的最小代價(jià)。同時(shí)設(shè)\(size[i]\)表示以\(i\)為根的子樹大小，那么容易得到以下狀態(tài)轉(zhuǎn)移方程：
\[ f[i][k]=min(f[i-size[j]][k]+w[i])_{j\in i} \]
但在實(shí)現(xiàn)時(shí)還有些細(xì)節(jié)需要注意。為了節(jié)約空間，可以在每一層遞歸中開一個(gè)數(shù)組記錄當(dāng)前的\(f\)的值，這樣就可以節(jié)約掉\(f\)的第二維。

這道題的毒瘤輸入也值得一提......

代碼

#include <iostream> #include <cstdio> #include <cstring> #include <map> #define N 402 using namespace std; int head[N],ver[N*2],nxt[N*2],l; int f[N],w[N],din[N],size[N],n,m,i,cnt,ans=0; char s[102]; string u,v; map<string,int> d; void insert(int x,int y) {l++;din[y]++;ver[l]=y;nxt[l]=head[x];head[x]=l; } void init(int x,int pre) {size[x]=1;for(int i=head[x];i;i=nxt[i]){int y=ver[i];if(y!=pre){init(y,x);size[x]+=size[y];}} } void dp(int x,int y) {int tmp[N]={0};for(int i=0;i<=n;i++) tmp[i]=f[i];for(int i=head[x];i;i=nxt[i]){int y=ver[i];dp(y,x);}for(int i=n;i>=size[x];i--){tmp[i]=min(tmp[i],tmp[i-size[x]]+w[x]);f[i]=min(f[i],tmp[i]);} } int main() {while(gets(s)){if(s[0]=='#') break;sscanf(s,"%d%d",&n,&m);memset(f,0x3f,sizeof(f));memset(head,0,sizeof(head));memset(din,0,sizeof(din));d.clear();f[0]=l=cnt=0;for(i=1;i<=n;i++){int p;cin>>u>>p;if(d[u]==0) d[u]=++cnt;w[d[u]]=p;char c=getchar();while(c==' '){cin>>v;if(d[v]==0) d[v]=++cnt;insert(d[u],d[v]);c=getchar();}}for(i=1;i<=n;i++){if(din[i]==0) init(i,0);}for(i=1;i<=n;i++){if(din[i]==0) dp(i,0);}ans=1<<30;for(i=m;i<=n;i++) ans=min(ans,f[i]);cout<<ans<<endl;}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/LSlzf/p/10973537.html

《新程序員》：云原生和全面數(shù)字化實(shí)踐50位技術(shù)專家共同創(chuàng)作，文字、視頻、音頻交互閱讀

總結(jié)

以上是生活随笔為你收集整理的[POJ 3345] Bribing FIPA的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。