【CF 702E】Analysis of Pathes in Functional Graph（倍增）

【CF 702E】Analysis of Pathes in Functional Graph（倍增）

题目大意：
n个点n条边的有向图，每个点有且只有一个后继。图中一定会有环。

问从每个点出发，走k条边，总花费，还有这段路程里的最小边权。

每个边每个点可以走多次，从每个点出发走k步为止。

k很大，第一反应是找环，找出所有环，然后想办法统计每个点到环的路长，然后……麻烦到吐血= =还搞不出来……

virtual后百度了一发，一看到倍增……厉害啊！！！大脑一阵嗡嗡声，太强了！

类似rmq的思想，一维表示区间长度，然后转移就行了。查询的时候把k拆成几个2的整数次幂，就完了。强强强强强！太妙了。

代码如下：

#include <iostream>#include <cmath>#include <vector>#include <cstdlib>#include <cstdio>#include <climits>#include <ctime>#include <cstring>#include <queue>#include <stack>#include <list>#include <algorithm>#include <map>#include <set>#define LL long long#define Pr pair<int,int>#define fread(ch) freopen(ch,"r",stdin)#define fwrite(ch) freopen(ch,"w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9+7;const double eps = 1e-8;const int maxn = 112345;struct Edge{    int v,mn;    LL sum;} rmq[40][maxn];LL k;int n;void init(){    for(int j = 1; (1LL<<j) <= k; ++j)        for(int i = 0; i < n; ++i)        {            rmq[j][i].sum = rmq[j-1][i].sum+rmq[j-1][rmq[j-1][i].v].sum;            rmq[j][i].mn = min(rmq[j-1][i].mn,rmq[j-1][rmq[j-1][i].v].mn);            rmq[j][i].v = rmq[j-1][rmq[j-1][i].v].v;            //printf("pow:%lld sum:%lld mn:%d en:%d\n",1LL<<j,rmq[j][i].sum,rmq[j][i].mn,rmq[j][i].v);        }}void query(int id){    LL tmp = k;    LL sum = 0;    int mn = 100000000;    for(int j = 0; (1LL<<j) <= tmp; ++j)        if((1LL<<j)&tmp)        {            tmp -= (1LL<<j);            sum += rmq[j][id].sum;            mn = min(mn,rmq[j][id].mn);            id = rmq[j][id].v;        }    printf("%lld %d\n",sum,mn);}int main(){    //fread("");    //fwrite("");    int u,v,w;    scanf("%d%lld",&n,&k);    for(int i = 0; i < n; ++i) scanf("%d",&rmq[0][i].v);    for(int i = 0; i < n; ++i)     {        scanf("%d",&rmq[0][i].mn);        rmq[0][i].sum = rmq[0][i].mn;    }    init();    for(int i = 0; i < n; ++i)        query(i);    return 0;}