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畅通project
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21271 Accepted Submission(s): 9154
经过调查评估。得到的统计表中列出了有可能建设公路的若干条道路的成本。
现请你编敲代码。计算出全省畅通须要的最低成本。
行相应村庄间道路的成本,每行给出一对正整数,各自是两个村庄的编号,以及此两村庄间道路的成本(也是正整数)。
为简单起见,村庄从1到M编号。当N为0时,所有输入结束。相应的结果不要输出。
若统计数据不足以保证畅通,则输出“?”。
3 3 1 2 1 1 3 2 2 3 4 1 3 2 3 2 0 100
3 ?AC-code:【Kruskal】#include<cstdio> #include<algorithm> using namespace std; int per[10005]; void pre() { for(int i=1;i<10005;i++) per[i]=i; } struct node { int b,e,v; }s[10000]; bool cmp(node a,node c) { return a.v<c.v; } int find(int a) { return a==per[a]?a:find(per[a]); } int join(int a,int c) { int fx=find(a); int fy=find(c); if(fy!=fx) { per[fy]=fx; return true; } return false; } int main() { int n,m,i,sum,ans; while(scanf("%d%d",&n,&m),n) { pre(); sum=0; for(i=0;i<n;i++) scanf("%d%d%d",&s[i].b,&s[i].e,&s[i].v); sort(s,s+n,cmp); for(i=0;i<n;i++) { if(join(s[i].b,s[i].e)) sum+=s[i].v; } ans=0; for(i=1;i<=m;i++) if(i==per[i]) ans++; if(ans>1) printf("?\n"); else printf("%d\n",sum); } return 0; }
【Prim】#include<cstdio> #include<cstring> #include<algorithm> #define max 0xffffff using namespace std; int main() { int n,m,i,j,a,b,c,ans,flag,k,min,s[105][105],dis[105],vis[105]; while(scanf("%d%d",&n,&m),n) { for(i=1;i<=m;i++) for(j=1;j<=m;j++) s[i][j]=max; while(n--) { scanf("%d%d%d",&a,&b,&c); s[a][b]=s[b][a]=c; } memset(vis,0,sizeof(vis)); for(i=2;i<=m;i++) dis[i]=s[1][i]; dis[1]=0; vis[1]=1; ans=flag=0; for(j=2;j<=m;j++) { min=max; for(i=2;i<=m;i++) if(!vis[i]&&dis[i]<min) { k=i; min=dis[i]; } if(min==max) { flag=1; break; } vis[k]=1; for(i=2;i<=m;i++) if(!vis[i]&&dis[i]>s[k][i]) dis[i]=s[k][i]; } if(flag) printf("?\n"); else { for(i=2;i<=m;i++) ans+=dis[i]; printf("%d\n",ans); } } return 0; }
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