Axis2 webservice 各种jar包使用(缺包的异常)
Constructing Roads
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12235 Accepted Submission(s): 4655
We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.
Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.
3 0 990 692 990 0 179 692 179 0 1 1 2
179
#include"stdio.h" #include"string.h" #include"math.h" #define N 105 const int Inf=10000; int main() { int n,q,i,j,a,b,ans,s,index,min; int map[N][N],mark[N],dis[N]; while(scanf("%d",&n)!=-1) { for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&map[i][j]); scanf("%d",&q); while(q--) //已经连通的的点距离改记为0; { scanf("%d%d",&a,&b); map[a][b]=0; map[b][a]=0; } s=1; //先加入一个点,然后更新其余点与该点的距离 ans=0; memset(mark,0,sizeof(mark)); mark[1]=1; for(i=1;i<=n;i++) dis[i]=map[s][i]; while(1) { index=1; min=Inf; for(i=1;i<=n;i++) if(!mark[i]&&dis[i]<min) //找到距离该点最近的点,加入集合 { index=i; min=dis[i]; } if(index==1) //完成建路任务 break; mark[index]=1; ans+=min; for(i=1;i<=n;i++) //若存在集合外的点距离新加入点的距离小于以前距离,更新 if(!mark[i]&&map[index][i]<dis[i]) dis[i]=map[index][i]; } printf("%d\n",ans); } return 0; }
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