PAT-Forwards on Weibo
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- 原题链接:Forwards on Weibo
- 题目大意:给你一个有向图、起点和最远能走的步数,让你计算一共可以经历多少个点
- 解法:无非就是遍历,给定起点,我们利用广度优先遍历的算法来做,使用queue来存储每一个节点,注意,在这里我们不仅仅要维护每个节点的坐标还要维护一个length值,代表的是从起点走到当前节点的路程距离,这个举例不能超过题中给出的约束条件。
- 代码如下:
#include <iostream>#include <cstring>#include <string>#include <queue>#include <cstdio>using namespace std;const int max_n = 1005;int relations[max_n][max_n];void display(pair<int, int> p){ cout<<p.first<<" "<<p.second<<endl;}int find_point_number(int s, int l, int n){ bool existed[max_n]; memset(existed, false, sizeof(existed)); queue<pair<int, int> > q; pair<int, int> p; p.first = s; p.second = 0; q.push(p); existed[s] = true; int res = 0; while(!q.empty()) { pair<int, int> curp = q.front(); //display(curp); q.pop(); if(curp.second >= l) { continue; } for(int i=1;i<=n;i++) { if(existed[i] == false && relations[curp.first][i] == 1) { existed[i] = true; //cout<<"find i is "<<i<<endl; res += 1; pair<int, int> added(i, curp.second+1); q.push(added); } } } return res;}int main(){ freopen("/home/give/PAT/ForwardsonWeibo.txt", "r", stdin); int n,l; cin>>n>>l; memset(relations, 0, sizeof(relations)); for(int i=1;i<=n;i++) { int num; cin>>num; for(int j=0;j<num;j++) { int index; cin>>index; relations[index][i] = 1; } } int points_num; cin>>points_num; for(int i=0;i<points_num;i++) { int point_index; cin>>point_index; cout<<find_point_number(point_index, l, n)<<endl; } return 0;}
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