FJNU摸底赛_acdream1681(bfs)

来源：互联网发布：郝斌c语言笔记编辑：程序博客网时间：2024/05/20 21:20

题意:

要从这边岸,踩着石头跳到那边岸;

给出对岸的距离y0,石头的数量n ,和一次最多跳多远d;

然后给出n块石头的横纵坐标;

问能不能跳到对岸,能输出YES和最少跳几次;

不能输出NO和离对岸最近的距离是多少;

思路:

bfs,先把第一次能跳到的放进队列,再把第二次能跳到的放进队列,一直找到最后;

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int N = 1005;struct point {int x,y;int step;}p[N];int n,y0,d;int getdis(point a, point b) {return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);}void bfs() {queue<point> q;int dis = y0;while(!q.empty())q.pop();for(int i = 0; i < n; i++) {if(p[i].y <= d) {dis = min(dis, y0 - p[i].y);p[i].step = 1;q.push(p[i]);}}while(!q.empty()) {point a = q.front();q.pop();if(y0 - a.y <= d) {printf("YES\n%d\n",a.step + 1);return;}for(int i = 0; i < n; i++) {if(getdis(a,p[i]) <= d * d && p[i].step == -1) {dis = min(dis, y0 - p[i].y);p[i].step = a.step + 1;q.push(p[i]);}}}printf("NO\n%d\n",dis);return;}int main() {int t;scanf("%d",&t);while(t--) {scanf("%d%d%d",&y0, &n, &d);for(int i = 0; i < n; i++) {scanf("%d%d",&p[i].x, &p[i].y);p[i].step = -1;}if(y0 <= d) {printf("YES\n1\n");}else {bfs();}}}

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