USACO Section 2.4 Bessie Come Home - 无限水法..Floyd模板题..囧..
2011-12-01 15:14
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Floyd模板水题..要注意的是因为给的边数显然是可能大于所有可能边数的..也就是两点间可能有多条路..在读入是判断下..再一个很重要!!..是无向边!!我就因为搞成有向边给WA了一次..
Program:
Program:
/* ID: zzyzzy12 LANG: C++ TASK: comehome */ #include<iostream> #include<istream> #include<stdio.h> #include<string.h> #include<math.h> #include<stack> #include<algorithm> #include<queue> using namespace std; int d[101][101],p,i,j,k; char c; int getdata() { c='1'; while (!(c>='A' && c<='Z') && !(c>='a' && c<='z')) c=getchar(); if (c>='A' && c<='Z') return c-'A'+1+26; return c-'a'+1; } void Floyd() { int i,j,k; for (k=1;k<=52;k++) for (i=1;i<=52;i++) for (j=1;j<=52;j++) if (d[i][j]-d[i][k]>d[k][j]) d[i][j]=d[i][k]+d[k][j]; } int main() { freopen("comehome.in","r",stdin); freopen("comehome.out","w",stdout); memset(d,0x7f,sizeof(d)); scanf("%d",&p); while (p--) { i=getdata(); j=getdata(); scanf("%d",&k); if (d[i][j]>k) d[i][j]=d[j][i]=k; } Floyd(); int ans=27; for (i=27;i<52;i++) if (d[i][52]<d[ans][52]) ans=i; printf("%c %d\n",'A'+ans-26-1,d[ans][52]); return 0; }
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