Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E: Bear with me, again..

There are bears everywhere! In fact, look around, there might be one around you!
Okay, maybe not.

Here's the story:
Larry is trying to come up with some kind of escape plan from the island.. and then realizes that he's struck with nothing except for Ryan, some duct tape, and some cardboard. What are the odds? He knows how to (somewhat) build a raft of sorts from it! Unforunately for Larry, since his craftmenship isn't quite good, not to mention he really have no tools, he can only build a raft that lasts K straight days at sea, after of which if it can get to an island, it can be fixed up and reused. He notices that he and Ryan can row at a speed of M miles per day. Still, with some luck, and Ryan's GPS (which came with the super-iPod, you see), he is able to see islands from far far away. Somehow all the islands are perfectly circular... and he knows exactly where home is.. can he get there, hopping on islands along the way?

So doing some calculations, he realizes his fate..

Input

Everything will in miles.
First line will contain the integers K and M, as defined above in the problem.
The second line of each input will contain three integers, space limited: x, y, r. x and y being the coordinates of the current island, the r being the radius of the island (somehow, all the islands are perfectly circular.)
Third line will contain three integers in the same format as above, will contain the location and radius of the island they want to get to.
Then there will be a single integer N which is followed by N lines, each in the format of x_i, y_i, and r_i, each representing an island coordinate and its radius in miles. N will be at most 100.

No islands will have overlaps, and all integers representing coordinates and radius will be less than 200.

Output

For each test case, output

Larry and Ryan will escape!

if they can get back to their home island in some way, or

Larry and Ryan will be eaten to death.

if they cannot get home.

Sample Input

1 1
0 0 1
0 3 1
0
1 1
0 0 1
0 5 1
0
1 1
0 0 1
0 6 1
1
0 3 1

Sample Output

Larry and Ryan will escape!
Larry and Ryan will be eaten to death.
Larry and Ryan will escape!

---

uHunt 說是 All pairs，但看懂題目就是 SSSP, Dijkstra 可做的。

簡單的說一天最長移動距離 K*M，他說了一次最多在海上航行 K 天，每天最多移動 M 距離。

之後把島之間算連線距離即可。

#include <stdio.h>
#include <math.h>
#define oo 1000000000
int main() {
    double K, M;
    double g[105][105];
    while(scanf("%lf %lf", &K, &M) == 2) {
        double mxDis = K*M, dis, tmp;
        double x[105], y[105], r[105];
        int n, i, j, k;
        scanf("%lf %lf %lf", &x[0], &y[0], &r[0]);
        scanf("%lf %lf %lf", &x[1], &y[1], &r[1]);
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%lf %lf %lf", &x[i+2], &y[i+2], &r[i+2]);
        n += 2;
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                tmp = sqrt(pow(x[i]-x[j],2) + pow(y[i]-y[j],2));
                if(tmp <= r[i]+r[j])
                    dis = 0;
                else
                    dis = tmp - r[i]- r[j];
                if(dis > mxDis) g[i][j] = oo;
                else g[i][j] = dis;
            }
        }
        for(k = 0; k < n; k++)
            for(i = 0; i < n; i++)
                for(j = 0; j < n; j++)
                    if(g[i][j] > g[i][k]+g[k][j])
                        g[i][j] = g[i][k]+g[k][j];
        if(g[0][1] != oo)
            puts("Larry and Ryan will escape!");
        else
            puts("Larry and Ryan will be eaten to death.");
    }
    return 0;
}