poj 2075(prim 算法)

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题目链接：点击打开链接‘

题目分析：最小生成树的小题目，适合练手

题目总结：

。。。小心啊写成“%1.f”一直出不来= =

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#define MAXN 1000#define INF 1000000000using namespace std;string data[MAXN];float grid[MAXN][MAXN],cable;//cable is the maxibool comp(string a,string b){    int len_a=a.length(),len_b=b.length();    if(len_a!=len_b) return false;    for(int i=0;i<len_a;i++)    {        if(a[i]!=b[i]) return false;    }    return true;}int find_loc(string a,int n){    for(int i=1;i<=n;i++)    {        if(1==operator==(a,data[i]))        return i;    }}float ans;void prim(int n){    int close[MAXN],i,u,num=1,merge;    float lowcost[MAXN],min_dis;    for(i=1;i<=n;i++)    {        lowcost[i]=INF,close[i]=0;    }    u=1;ans=0;close[1]=-1;    while(1)    {        if(num==n)          break;        min_dis=INF;        for(u =1;u<=n;u++)//多加了一成循环在【u】中找            {            if(close[u]==-1)            {            for(i=1;i<=n;i++)            {            if(close[i]!=-1&&grid[u][i]+1.0>1e-8)//判断是否有边            {                double tmp=grid[u][i];                if(tmp<lowcost[i])                lowcost[i]=tmp;                 if(lowcost[i]<min_dis)                 {                     merge=i;                     min_dis=lowcost[i];                 }            }            }            }        }        ans+=min_dis;close[merge]=-1;       num++;    }}int main(){    freopen("in.txt","r",stdin);    int n,m,i,a,b;    string x,y;    float dis;    while(scanf("%f",&cable)!=EOF)    {        scanf("%d",&n);        getchar();        for(i=1;i<=n;i++)            getline(cin,data[i]);        scanf("%d",&m);        for(i=1;i<=n;i++)     //用memset 的效果一样          for(int j=1;j<=n;j++)            grid[i][j]=-1.0;        for(i=1;i<=m;i++)        {            getchar();            getline(cin,x,' ');//c++对getline不接受，会ce            getline(cin,y,' ');            scanf("%f",&dis);            a = find_loc(x,n); b = find_loc(y,n);            grid[a][b]=grid[b][a]=dis;        }        prim(n);        if(ans>cable) printf("Not enough cable\n");        else printf("Need %.1f miles of cable\n",ans);//。。。小心啊写成“%1.f”一直出不来= =    }    return 0;}