zoukankan      html  css  js  c++  java
  • Ice_cream’s world III(prime)

    Ice_cream’s world III

    Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 6   Accepted Submission(s) : 3
    Problem Description
    ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world; the queen’s requirement is like II problem, beautifies the roads, by which there are some ways from every city to the capital. The project’s cost should be as less as better.
     
    Input
    Every case have two integers N and M (N<=1000, M<=10000) meaning N cities and M roads, the cities numbered 0…N-1, following N lines, each line contain three integers S, T and C, meaning S connected with T have a road will cost C.
     
    Output
    If Wiskey can’t satisfy the queen’s requirement, you must be output “impossible”, otherwise, print the minimum cost in this project. After every case print one blank.
     
    Sample Input
    2 1 0 1 10 4 0
     
    Sample Output
    10 impossible
    题解:
    联通所有的路的最小代价;不连通就输出impossible
    prime代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 const int INF=0x3f3f3f3f;
     4 const int MAXN=1010;
     5 int N,M,answer;
     6 int map[MAXN][MAXN],low[MAXN],vis[MAXN];
     7 void prime(){
     8         int k,flot=1,temp;
     9         memset(vis,0,sizeof(vis));
    10         vis[0]=1;
    11         for(int i=0;i<N;i++)
    12             low[i]=map[0][i];
    13         for(int i=0;i<N;i++){
    14             temp=INF;
    15             for(int j=0;j<N;j++)
    16                 if(!vis[j]&&temp>low[j])temp=low[k=j];
    17             if(temp==INF){
    18                 if(flot==N)printf("%d
    ",answer);
    19                 else puts("impossible");
    20                 break;
    21             }
    22             answer+=temp;
    23             vis[k]=1;
    24             flot++;
    25             for(int j=0;j<N;j++)
    26                 if(!vis[j]&&low[j]>map[k][j])
    27                 low[j]=map[k][j];
    28         }
    29 }
    30 int main(){
    31     int a,b,c;
    32     while(~scanf("%d%d",&N,&M)){answer=0;
    33         memset(map,INF,sizeof(map));
    34         while(M--){
    35             scanf("%d%d%d",&a,&b,&c);
    36             if(c<map[a][b])
    37                 map[a][b]=map[b][a]=c;
    38         }
    39         prime();
    40         puts("");
    41     }
    42 return 0;
    43 }
  • 相关阅读:
    xml传数据
    简单实用的GroupBox控件
    漂亮的NavMenu导航控件
    使用设计模式构建通用数据库访问类
    Windows路由表详解
    zz Linux Shell常用技巧(目录)
    Ubuntu Linux 环境变量PATH设置
    zz eclipse.ini内存设置
    find 用法
    zz【java规范】Java spi机制浅谈
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4725313.html
Copyright © 2011-2022 走看看