題干：

Jungle Roads
Time Limit: 2000/1000 MS (Java/Others) ? ?Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5505 ? ?Accepted Submission(s): 3976
Problem Description

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.?


The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.?

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The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.?

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Sample Input
9
A ?2 ?B ?12 ?I ?25
B ?3 ?C ?10 ?H ?40 ?I ?8
C ?2 ?D ?18 ?G ?55
D ?1 ?E ?44
E ?2 ?F ?60 ?G ?38
F ?0
G ?1 ?H ?35
H ?1 ?I ?35
3
A ?2 ?B ?10 ?C ?40
B ?1 ?C ?20
0
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Sample Output
216

30

解題報(bào)告：

? ? ? ? ? ? 最小生成樹(shù)再來(lái)一發(fā)。

ac代碼：

#include<cstdio> #include<algorithm> #include<iostream> using namespace std; struct Node {int u,v;int w;bool operator< ( const Node & b) const{return w<b.w;} } node[105];int f[100],n,cur,cnt,ans; void init() {cnt = 0;cur = 0;ans = 0;for (int i = 0; i <= 27; i++) {f[i] = i; } } int getf(int u) {return u==f[u] ? u : f[u]=getf(f[u]); } void merge(int u,int v) {int t1,t2;t1=getf(u);t2=getf(v);if(t1!=t2) f[t2]=t1; } void input() {char ch[5];int a;for (int i = 1; i <= n - 1; i++) {scanf("%s %d", ch, &a);int u = ch[0] - 'A' + 1;for (int j = 1; j <= a; j++) {int b;scanf(" %s %d", ch, &b);int v = ch[0] - 'A' + 1;node[cnt].u = u;node[cnt].v = v;node[cnt].w = b; cnt++;}} }int main() {while (scanf("%d", &n) ) {getchar();//如果不用getchar 那input中就用%s別用%c if (n==0) break;init();input();sort(node,node+cnt);for(int i = 0; i<cnt; i++) {if(getf(node[i].u) != getf(node[i].v) ) {merge(node[i].u,node[i].v);ans+=node[i].w;cur++;}if(cur==n-1) {break;}}printf("%d\n", ans);}return 0; }

貼克魯斯特爾算法ac代碼：（常用！ac代碼1就是用此法）

//hdu-1301-Jungle Roads(克魯斯卡爾) //題目大意：給出個(gè)村莊之間的道路及其維修費(fèi)；從中選取一些路使道路的維修費(fèi)最小且保證各村莊 //之間道路暢通； //解題思路： //本題最核心的還是構(gòu)圖，由于輸入的村莊用字母代替這就需要在構(gòu)圖中將字母替換成數(shù)字依次給每個(gè)村莊編號(hào)； //需要注意的由于輸入的有字符型數(shù)據(jù)，這就要加一些 getchar()來(lái)吸收掉中間的換行空格啥的； //初始化 per[n] 數(shù)組一定要注意，把所有村莊都初始化就行啦； //具體如下： #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int per[30]; int n,m; struct fun{ //定義結(jié)構(gòu)體數(shù)組road[100],用來(lái)存放起點(diǎn)、終點(diǎn)和距離 int s;int e;int p; }road[100]; int cmp(fun x,fun y) {return x.p < y.p; } void init() //初始化 per 數(shù)組 ；注意初始完 n個(gè)村莊即可； {int d;for(d=1;d < = n;d++)per[d]=d; } int find(int x) {int r=x;while(r!=per[r])r=per[r];return r; } bool link(int x,int y) {int fx=find(x),fy=find(y);if(fx!=fy){per[fx]=fy;return true;}return false; } int main() {int j,k,l,sum,i,cnt,a1,a2;char str1,str2;while(scanf("%d",&n)&&(n!=0)){getchar();memset(road,0,sizeof(road));memset(per,0,sizeof(per));init();for(i=1,cnt=1;i < n;i++) //輸入數(shù)據(jù)，并構(gòu)圖； {scanf("%c%d",&str1,&m);for(j=1;j<=m;j++){getchar();scanf("%c%d",&str2,&l);road[cnt].s=str1-'A'+1; //將字母編號(hào)轉(zhuǎn)化為數(shù)字編號(hào)并存入結(jié)構(gòu)體數(shù)組中 road[cnt].e=str2-'A'+1;road[cnt].p=l;cnt++;}getchar();} //構(gòu)圖完成，剩下的就是克魯斯卡爾啦； sort(road,road+cnt,cmp);sum=0;for(j=1;j<=cnt;j++){if(link(road[j].s,road[j].e))sum+=road[j].p;}printf("%d\n",sum);}return 0; }

貼Prim算法ac代碼：

//hdu-1301(普里姆算法) //本題用普里姆算法求最小生成樹(shù)直接用普里姆的思想構(gòu)造函數(shù)即可； //需要注意的是本題再輸入數(shù)據(jù)的是時(shí)候要先對(duì) map[1000][1000]中的前 n*n //個(gè)元素初始化為 一個(gè)與 min相同或大于它的數(shù)；例如，0xfffffff即可； //(n是結(jié)點(diǎn)的個(gè)數(shù)，由于需要在 n*n 大小的矩陣中構(gòu)圖，所以初始化 n*n 就行)； //為什么要對(duì) map數(shù)組初始化呢？一是由于在prim函數(shù)中 map要對(duì)s[1000]數(shù)組賦初值；如果不對(duì)map初始化； //那么，s[]數(shù)組中沒(méi)有數(shù)據(jù)的就一直為零，與下面的 min比較時(shí)，min是永遠(yuǎn)大于它們的；這樣得到的min就一直為零 //最終導(dǎo)致sum為零；還有，不對(duì)map按上述初始化，在更新時(shí)候也無(wú)法完成 ； #include<stdio.h> #include<string.h> #define N 0xfffff int map[1000][1000]; int n,m,sum,cnt; void input() {char c1,c2;int i,j,x,y,price;cnt=0;for(i=1;i<=n;i++) //對(duì) map數(shù)組初始化為 N ;{for(j=1;j<=n;j++)map[i][j]=N;}for(i=1;i<n;i++) //輸入數(shù)據(jù)，并構(gòu)圖； {scanf("%c%d",&c1,&m);x=c1-'A'+1;for(j=1;j<=m;j++){getchar();scanf("%c%d",&c2,&price);y=c2-'A'+1;map[x][y]=map[y][x]=price;}getchar();} } void prim() //普里姆； {int i,j,k,min,v,a,l;int s[1000],vis[1000];memset(vis,0,sizeof(vis));for(i=1;i<=n;i++) s[i]=map[1][i]; //初始化 s 數(shù)組； s[1]=0; //本身到本身的距離為零 vis[1]=1; //標(biāo)記第一個(gè)點(diǎn)； sum=0;for(v=1;v<n;v++){min=N;k=1;for(j=1;j<=n;j++)if(!vis[j]&&min>s[j]){min=s[j]; //尋找當(dāng)前最小權(quán)值； k=j;}vis[k]=1; //標(biāo)記頂點(diǎn)k,表示k已連上樹(shù)； sum+=min; //將當(dāng)前最小權(quán)值累加到sum中； for(a=1;a<=n;a++){if(!vis[a]&&s[a]>map[k][a]) //標(biāo)記當(dāng)前最小權(quán)值的頂點(diǎn)后，更新s[] 數(shù)組中的權(quán)值 ； s[a]=map[k][a];}}printf("%d\n",sum); } int main() {char c1,c2;int i,j,x,y,price;while(scanf("%d",&n)&&(n!=0)){getchar();input();prim();}return 0; }

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總結(jié)：

? ? ? 有的時(shí)候用%s讀字符，不是因?yàn)閯e的，就是因?yàn)楹锰幚砜崭?#xff0c;回車這些煩人的東西！！這件事情告訴我們，少用%c多用%s，少用char多用char[ ] ！！

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