标签:
bzoj1624[Usaco2008 Open] Clear And Present Danger 寻宝之路
题意:
求按点1-a1-a2...-an-n走的最短路长度是多少。点数小于等于100。
题解:
floyd。
代码:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define maxn 110 5 #define inc(i,j,k) for(int i=j;i<=k;i++) 6 using namespace std; 7 8 inline int read(){ 9 char ch=getchar(); int f=1,x=0; 10 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1; ch=getchar();} 11 while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘,ch=getchar(); 12 return f*x; 13 } 14 int map[maxn][maxn],a[maxn*100],n,m,ans; 15 int main(){ 16 n=read(); m=read(); inc(i,1,m)a[i]=read(); inc(i,1,n)inc(j,1,n)map[i][j]=read(); 17 inc(k,1,n)inc(i,1,n)inc(j,1,n)if(map[i][k]+map[k][j]<map[i][j])map[i][j]=map[i][k]+map[k][j]; 18 ans+=map[1][a[1]]; inc(i,2,m)ans+=map[a[i-1]][a[i]]; ans+=map[a[m]][n]; printf("%d",ans); return 0; 19 }
20160804
bzoj1624[Usaco2008 Open] Clear And Present Danger 寻宝之路*
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原文地址:http://www.cnblogs.com/YuanZiming/p/5743987.html