poj-2253

// 6048K   547MS   Java    
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import java.util.Queue;
import java.util.ArrayList;
import java.util.LinkedList;
import java.math.*;

public class Main{
    private static final long UNREACH = -1;
    private static final int MAX = 210;
    private static int sStoneNum;
    private static double[] dis;
    private static boolean[] djFlag;

    public static double[][] stoneDistance;
    public static Stone[] stones;

    public class Stone {
        public int x;
        public int y;
    }

    public void reset() {
        sStoneNum = 0;
        for (int i = 0; i < MAX; i++) {
            dis[i] = 0;
            djFlag[i] = false;
            for (int j = 0; j < MAX; j++) {
                stoneDistance[i][j] = UNREACH;
            }
        }
    }

    public double dj(int stoneNum) {
        for (int i = 0; i < stoneNum; i++) { // every node is connected, no need to check if connected
            dis[i] = stoneDistance[0][i];
        }
        djFlag[0] = true; // begin stone join V1;

        for (int j = 1; j < sStoneNum; j++) { // need sStoneNum-1 time caculate
            int minId = -1;
            int i = 0;
            double minRange = 99999999;
            for (; i < sStoneNum; i++) {
                if (!djFlag[i]) {
                    if (minRange > dis[i]) {
                        minId = i;
                        minRange = dis[i];
                    }
                }
            }

            djFlag[minId] = true; // minId node join V1

            for (i = 0; i < sStoneNum; i++) { // update other nodes not in V1
                if (!djFlag[i]) {
                    double maxRange = minRange > stoneDistance[minId][i] ? minRange : stoneDistance[minId][i];
                    dis[i] = dis[i] < maxRange ? dis[i] : maxRange;
                }
            }
        }
        return dis[1]; // floya is in 2th stone
    }

    public void solve(long caseId) {
        System.out.println("Scenario #" + caseId);
        // for (int i = 0; i < sStoneNum; i++) {
        //     System.out.println(stoneDistance[0][i]);
        // }
        String minRange = String.format("%.3f", dj(sStoneNum));
        // double minRange = dj(sStoneNum);
        System.out.println("Frog Distance = " + minRange + "\n");
        reset();
    }

    public Main() {
        
        stones = new Stone[MAX];
        for (int i = 0; i < MAX; i++) {
            stones[i] = new Stone();
        }
        dis = new double[MAX];
        stoneDistance = new double[MAX][MAX];
        djFlag = new boolean[MAX];
        reset();
    }

    public static double getDistance(int x1, int x2, int y1, int y2) {
        int xD = x1 - x2;
        int yD = y1 - y2;
        double orginalRes = Math.sqrt(xD*xD + yD*yD);
        BigDecimal tmp = new BigDecimal(orginalRes);
        return tmp.setScale(3, BigDecimal.ROUND_HALF_UP).doubleValue();
    }

    public static void main(String args[]) {
        Main solver = new Main();
        Scanner scanner = new Scanner(System.in);
        int caseId = 1;
        while(true) {
            sStoneNum = scanner.nextInt();
            if (sStoneNum == 0) {
                return;
            }
            
            for (int i = 0; i < sStoneNum; i++) {
                int x = scanner.nextInt();
                int y = scanner.nextInt();
                stones[i].x = x;
                stones[i].y = y;
                stoneDistance[i][i] = 0;
                for (int j = 0; j < i; j++) {
                    double distance = getDistance(x, stones[j].x, y, stones[j].y);
                    // System.out.println(distance);
                    stoneDistance[i][j] = distance;
                    stoneDistance[j][i] = distance;
                }
            }
            solver.solve(caseId++);
        }
        
    }

}

6048K   547MS   Java   
和1797一个模子，就是大小关系变了一下罢了，简直是做一送一，不过还是因为输出忘了加描述WA了一次 囧。

不同的是，这次的图是全直通图，因为题目没有规定青蛙最长能跳多远，所以默认青蛙可以从一个stone跳到任何其他的stone，

化成图就是图的每一个node都是互相直接连通的，因此在dj算法时，不用考虑两点之间是否直接连通。

麻烦一点的就是要先把每个stone之间的距离算出来，也简单，边接受输入边算就行了，注意保留3位四舍五入小数.

至于其他的步骤，和1797完全一样，

先解释一下题，青蛙在从起点跳到终点的path中，最长的一跳被成为 frogRange， 而因为从起点到终点会有多条不同的path，

那么要求就是这些path中最小的frogRange。

证明也一样：

假定现在从其实stoneB 到其他的没有确定到B的frogRange的其他stone， S1...SN,  而R1.....RN(目前从1...N点到B的最大frogRange)

满足R1

反证，如果存在其他的从stone1到B的path，那么该path必定会经过其他的stone， R2....RN, 而这些stone到B的frogRange都比R1大，

那么最终的frogRange也一定比R1大，因为按照题意，frograng是取path中最长的一跳.