Gym 100851F Froggy Ford(dijkstra)

题意

有条宽为w的河流，两岸分别在x = 0, x = w处，河中间有n个石板。在河的一岸有一只青蛙想通过石板跳到对岸去。现在可以在河
中间某个位置多加一块石板，使得在单步跳跃中的最大值最小。

思路

Dijkstra变形，用两维来表示是否路径中使用过额外的石板。dis[0][u]表示没使用额外石板的最大最小，dis[1][u]表示使
用额外石板的最大最小。然后就用Dijkstra格式进行转移了。

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2016
File Name   :
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
#define MEM(x,y) memset(x, y,sizeof x)
#define pk push_back
#define lson rt << 1
#define rson rt << 1 | 1
#define bug cout << "BUG HERE\n"
#define ALL(v) (v).begin(), (v).end()
#define lowbit(x) ((x)&(-x))
#define Unique(x) sort(ALL(x)); (x).resize(unique(ALL(x)) - (x).begin())
#define BitOne(x) __builtin_popcount(x)
#define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
#define Rep(i, l, r) for (int i = l;i <= r;++i)
#define Rrep(i, r, l) for (int i = r;i >= l;--i)
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> ii;
typedef pair<ii,int> iii;
const double eps = 1e-8;
const double pi = 4 * atan(1);
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int nCase = 0;
const int maxn = 1e3 + 123;
struct point {double x, y;void read() {scanf("%lf%lf", &x, &y);}double operator ^ (const point& rhs) {return sqrt((x - rhs.x) * (x - rhs.x) + (y - rhs.y) * (y - rhs.y));}
}a[maxn];
struct Edge {int v, nxt;double dis;
}e[maxn*maxn*4];
int head[maxn], ecnt;
void inline addedge(int u,int v,double dis) {e[ecnt] = Edge{v, head[u], dis}, head[u] = ecnt++;e[ecnt] = Edge{u, head[v], dis}, head[v] = ecnt++;
}
double dis[2][maxn];
int n, w;
struct node {int u;double md;bool used;point o;bool operator < (const node& rhs) const {return md > rhs.md || (md == rhs.md && used > rhs.used);// return md > rhs.md;}
};
void gao() {point ans;double m = INF;priority_queue<node> que;Rep(i, 0, n + 1) dis[0][i] = dis[1][i] = INF;que.push(node{0, 0, false, point{0, 0}});dis[0][0] = 0;while(!que.empty()) {node temp = que.top();que.pop();int u = temp.u;if (u == n + 1) {printf("%.12lf %.12lf\n", temp.o.x, temp.o.y);return ;// if (temp.md < m) {//     m = temp.md;//     ans.x = temp.o.x;//     ans.y = temp.o.y;// }continue;}for (int i = head[u]; ~i;i = e[i].nxt) {int v = e[i].v;if (v == 0) continue;if (temp.used) {if (dis[1][v] > max(temp.md, e[i].dis)) {dis[1][v] = max(temp.md, e[i].dis);que.push(node{v, dis[1][v], true, temp.o});}}else {double x, y;//边界处理，容易错if (u != 0 && v != n + 1) {x = (a[u].x + a[v].x) / 2.0;y = (a[u].y + a[v].y) / 2.0;}if (u != 0 && v == n + 1) {x = (a[u].x + w) / 2.0;y = a[u].y;}if (u == 0 && v == n + 1) {x = w / 2.0;y = 0;}if (u == 0 && v != n + 1) {x = a[v].x / 2.0;y = a[v].y;}// printf("[%.2lf, %.2lf]\n", x, y);if (dis[0][v] > max(temp.md, e[i].dis)) {dis[0][v] = max(temp.md, e[i].dis);que.push(node{v, dis[0][v], false, point{0, 0}});}if (dis[1][v] > max(temp.md, e[i].dis / 2.0)) {dis[1][v] = max(temp.md, e[i].dis / 2.0);que.push(node{v, dis[1][v], true, point{x, y}});}}}}printf("%.12lf %.12lf\n", ans.x, ans.y);
}
int main(int argc, const char * argv[])
{    // freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);freopen("froggy.in","r",stdin);freopen("froggy.out","w",stdout);while(~scanf("%d%d", &w, &n)) {Rep(i, 1, n) a[i].read();memset(head, -1, sizeof head), ecnt = 0;int vs = 0, vt = n + 1;addedge(vs, vt, w);Rep(i, 1, n) {addedge(vs, i, a[i].x);addedge(i, vt, w - a[i].x);Rep(j, i + 1, n) {addedge(i, j, a[i] ^ a[j]);}}if (n == 0) {printf("%.12lf 0\n", w / 2.0);continue;}gao();}// showtime;return 0;
}

Gym 100851F Froggy Ford(dijkstra)相关推荐

Gym - 100851F Froggy Ford (Dijkstra)
题目链接:https://cn.vjudge.net/problem/Gym-100851F 这道题是POJ-2253的升级版:https://blog.csdn.net/GYH0730/articl ...
Gym - 100851F Froggy Ford kruskal
题目链接: http://acm.hust.edu.cn/vjudge/problem/307216 Froggy Ford Time Limit: 3000MS 题意青蛙过河,河中有若干个石头,现 ...
Gym - 100851F - Froggy Ford（dijkstra）
题目链接参考 http://blog.csdn.net/KIJamesQi/article/details/52214990 题意蛤蛤要从这岸去到对岸,河中有n块石头,现可以在河中添加一块石头 ...
Gym - 100851F Froggy Ford
题目链接: http://codeforces.com/gym/100851/attachments 题解: 二分枚举长度,BFS暴力找路径,注意边界的处理代码: #include <set& ...
【二分答案】【DFS】【分类讨论】Gym - 100851F - Froggy Ford
题意:河里有n块石头,一只青蛙要从左岸跳到右岸,你可以再在任意一个位置放一块石头,使得在最优方案下,青蛙单步跳的距离的最大值最小化,输出该位置. 将原图视作完全图,二分答案mid,然后在图中只保留小于 ...
Froggy Ford Gym - 100851F
http://codeforces.com/gym/100851/attachments 青蛙过河 n个石头当作n个点任意两点之间都有边边的花费即为两点欧式距离现在可以在任意位置加一个石头使 ...
【最短路】NEERC15 F Froggy Ford(2015-2016 ACM-ICPC)(Codeforces GYM 100851)
题目链接: http://codeforces.com/gym/100851 题目大意: 一只青蛙跳过宽为W的河,河中游N个石头,坐标xi,yi,现在往河中间添加一个石头,使得每次跳跃的最大的距离最小 ...
Gym-100851F Froggy Ford 最短路变形 dijkstra || spfa
题意:有条宽为w的河流,两岸分别在x = 0, x = w处,河中间有n个石头.在河的左岸有一只青蛙想通过石头跳到对岸去.现在可以在河中间某个位置多加一块石头,使得青蛙在单步跳跃中的最大值最小,问应在 ...
【最短路】NEERC15 F Froggy Ford (Codeforces GYM 100851)
题目链接: http://codeforces.com/gym/100851 题目大意: 一只青蛙跳过宽为W的河,河中游N个石头,坐标xi,yi,现在往河中间添加一个石头,使得每次跳跃的最大的距离最小 ...

Gym 100851F Froggy Ford(dijkstra)

题意

思路

Gym 100851F Froggy Ford(dijkstra)相关推荐

最新文章

热门文章