http://codeforces.com/contest/702
题意：n个点，n条边，每个点出边只有一条，问从每个点出发经过k条边的边权和，以及边权最小值
思路：
f[i][j] 第i个点出发，经过2^j条边后的相连点 其余类似
二进制巧妙解决问题应用太广了
1 // #pragma comment(linker, "/STACK:102c000000,102c000000") 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <sstream> 6 #include <string> 7 #include <algorithm> 8 #include <list> 9 #include <map> 10 #include <vector> 11 #include <queue> 12 #include <stack> 13 #include <cmath> 14 #include <cstdlib> 15 // #include <conio.h> 16 using namespace std; 17 #define pi acos(-1.0) 18 const int N = 1e5+10; 19 const int MOD = 1e9+7; 20 #define inf 0x7fffffff 21 typedef long long LL; 22 23 void frein(){freopen("in.txt","r",stdin);} 24 void freout(){freopen("out.txt","w",stdout);} 25 inline LL read(){LL x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}return x*f;} 26 const int M =35; 27 28 int f[N][M]; 29 LL sum[N][M]; 30 int mn[N][M]; 31 int main(){ 32 int n; 33 LL k; 34 n=read(),k=read(); 35 for(int i=0;i<n;i++) scanf("%d",&f[i][0]); 36 for(int i=0;i<n;i++) {scanf("%d",&sum[i][0]);mn[i][0]=sum[i][0];} 37 for(int j=1;j<M;j++){ 38 for(int i=0;i<n;i++){ 39 f[i][j]=f[f[i][j-1]][j-1]; 40 sum[i][j]=sum[f[i][j-1]][j-1]+sum[i][j-1]; 41 mn[i][j]=min(mn[i][j-1],mn[f[i][j-1]][j-1]); 42 } 43 } 44 LL ans; 45 for(int i=0;i<n;i++){ 46 int v=i,minn=inf; 47 LL K=k; 48 ans=0; 49 for(int j=M-1;j>=0;j--){ 50 if((1LL<<j)<=K){ 51 ans+=sum[v][j]; 52 minn=min(minn,mn[v][j]); 53 v=f[v][j]; 54 K-=(1LL<<j); 55 } 56 } 57 printf("%I64d %d\n",ans,minn); 58 } 59 return 0; 60 }