Poj_2253 Frogger(最短路-SPFA+Dijkstra)

题意：

青蛙1想要到青蛙2那儿去，中间有多个石头可供跳跃。求The frog distance.

The frog distance: the minimum necessary jump range over all possible paths between the two stones.

也就是说求所有路径中必须弹跳距离最短的那个，必须弹跳距离由路径中距离最远的那两个石头决定。

思路：

最开始题意理解了半天。。不是传统的最短路，从起点到终点路径的总长度不要求，但是对青蛙跳跃能力（最大跳跃距离）的要求尽可能小。也就是说青蛙跳多少次没关系，但是在这个过程中青蛙都不用跳的特别远。

因为仍然要保证起点和终点的可达，所以仍用最短路来做。但是dis数组的定义要修改。

定义如下：

dis[i]:起点到i的所有路径中最小的必须弹跳距离；

对应的状态改变方程：

dis[i] = min(dis[i],max(dis[tmp],edge[tmp][i]);

代码实现：

//SPFA
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAX_V = 210;
const int INF = 0x7ffffff;

struct Node{
int x,y;
};

int N;
int cnt;
queue<int> que;
Node node[MAX_V];
bool inque[MAX_V];
double dis[MAX_V];
double edge[MAX_V][MAX_V];
int main(){
cnt = 1;
while( scanf("%d",&N) != EOF ){
if( N == 0 ){
break;
}
for( int i = 1; i <= N; i++ ){
scanf("%d%d",&node[i].x,&node[i].y);
}
for( int i = 1; i <= N; i++ ){
for( int j = i+1; j <= N; j++ ){
int dx = node[i].x-node[j].x;
int dy = node[i].y-node[j].y;
double cost = sqrt(dx*dx+dy*dy);
edge[i][j] = cost;
edge[j][i] = cost;
}
dis[i] = INF;
edge[i][i] = 0;
}
memset(inque,false,sizeof(inque));
dis[1] = 0;
que.push(1);
inque[1] = true;
while( !que.empty() ){
int tmp = que.front();
que.pop();
inque[tmp] = false;
for( int i = 1; i <= N; i++ ){
if( dis[i] > max(dis[tmp],edge[tmp][i]) ){
dis[i] = max(dis[tmp],edge[tmp][i]);
if( inque[i] == false ){
que.push(i);
inque[i] = true;
}
}
}
}
printf("Scenario #%d\n",cnt++);
printf("Frog Distance = %.3f\n\n",dis[2]);
}
return 0;
}

//Dijkstra
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAX_V = 210;
const int INF = 0x7ffffff;

struct Node{
int x,y;
};

struct Pos{
int to;
double dis;
Pos(const int to,const double dis){
this->to = to;
this->dis = dis;
}
};

int N;
int cnt;
queue<Pos> que;
Node node[MAX_V];
double dis[MAX_V];
double edge[MAX_V][MAX_V];
int main(){
cnt = 1;
while( scanf("%d",&N) != EOF ){
if( N == 0 ){
break;
}
for( int i = 1; i <= N; i++ ){
scanf("%d%d",&node[i].x,&node[i].y);
}
for( int i = 1; i <= N; i++ ){
for( int j = i+1; j <= N; j++ ){
int dx = node[i].x-node[j].x;
int dy = node[i].y-node[j].y;
double cost = sqrt(dx*dx+dy*dy);
edge[i][j] = cost;
edge[j][i] = cost;
}
dis[i] = INF;
edge[i][i] = 0;
}
dis[1] = 0;
que.push(Pos(1,0));
while( !que.empty() ){
Pos tmp = que.front();
que.pop();
if( dis[tmp.to] < tmp.dis ){
continue;
}
for( int i = 1; i <= N; i++ ){
if( dis[i] > max(tmp.dis,edge[tmp.to][i]) ){
dis[i] = max(tmp.dis,edge[tmp.to][i]);
que.push(Pos(i,dis[i]));
}
}
}
printf("Scenario #%d\n",cnt++);
printf("Frog Distance = %.3f\n\n",dis[2]);
}
return 0;
}