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

Oh No!!! The predator has entered the room again. But thistime it is a different kind of room.

The room is a square of size10000 X 10000. There are several compartments of different shapes and sizes,situated strictly inside the room. The walls of different compartments don’toverlap, but a compartment can be completely inside that of another.

This time the predator haslearned to hop over walls, but it can jump over at most one wall at a time.Given the starting and ending coordinates of the predator, and the positions ofthe compartments, your job is to find out the minimum number of hops requiredfor the predator to reach his destination from the start.

The starting and ending positionsare never on the boundary of any compartment.

Input

The first line of input is aninteger (T ≤ 20), that indicates the number of test cases. Eachcase starts with an integer

(n ≤ 20), thatdetermines the number of compartments in the room. The next n lines give the positions of thecompartments. The compartments are simple polygons. The description of eachcompartment starts with an integers (S ≤ 10), that gives the numberof sides of the polygon, followed by S pairsof x, y coordinates in order. Nextthere is an integer Q thatdetermines the number of queries for this scenario. Each of the next Q lines contains 4 integers x1, y1, x2, y2. (x1, y1) is the starting position and (x2, y2) is the ending position.

The lower left and upper rightcoordinates of the room are (0, 0) and (10000, 10000) respectively.

Output

For each case, output the casenumber. Then for each query, output the minimum number of hops required.

ProblemSetter: Sohel Hafiz

Next Generation Contest 2

Illustration

The following diagram depicts the first sample input.

Pink spot à starting position

Black spot à target

The three hops are shown by three broad line segments.

利用射線法 - 判斷一個點是不是在一個簡單多邊形內。
判斷的方法 從那個點拉一條水平射線, 求交到多邊形的點個數。
偶數在多邊形外，奇數則在多邊內。 (核心代碼來自 DJWS)

然後利用這個方法，建出這棵關係樹。
然後求最短路徑。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <vector>
using namespace std;
vector<int> g[30];
int bg[30][30];
int used[30];
struct Pt {
    double x, y;
};
struct Polygon {
    Pt p[20];
    int n;
};
Polygon ply[30];
int parent[30];
int inPolygon(Pt p[], Pt q, int n) {
    int i, j, k = 0;
    for(i = 0, j = n-1; i < n; j = i++) {
        if(p[i].y > q.y != p[j].y > q.y &&
           q.x < (p[j].x-p[i].x)*(q.y-p[i].y)/(p[j].y-p[i].y)+p[i].x)
           k++;
    }
    return k&1;
}
void build(int nd, int n) {
    int i, j;
    for(i = 0; i <= n; i++) {
        if(used[i] == 0 && inPolygon(ply[nd].p, ply[i].p[0], ply[nd].n)) {
            for(j = 0; j <= n; j++)
                if(i != j && used[j] == 0 && inPolygon(ply[j].p, ply[i].p[0], ply[j].n))
                    break;
            if(j == n+1) {
                used[i] = 1;
                parent[i] = nd;
                g[nd].push_back(i);
                bg[i][nd] = bg[nd][i] = 1;
                build(i, n);
            }
        }
    }
}
int treefind(int nd, Pt p) {
    for(vector<int>::iterator it = g[nd].begin();
        it != g[nd].end(); it++) {
        if(inPolygon(ply[*it].p, p, ply[*it].n)) {
            return treefind(*it, p);
        }
    }
    return nd;
}
int main() {
    int t, n, cases = 0;
    int i, j, k;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++) {
            scanf("%d", &ply[i].n);
            for(j = 0; j < ply[i].n; j++)
                scanf("%lf %lf", &ply[i].p[j].x, &ply[i].p[j].y);
        }
        ply[i].n = 4;
        ply[i].p[0].x = -1, ply[i].p[0].y = -1;
        ply[i].p[1].x = -1, ply[i].p[1].y = 10005;
        ply[i].p[2].x = 10005, ply[i].p[2].y = 10005;
        ply[i].p[3].x = 10005, ply[i].p[3].y = -1;
        for(i = 0; i <= n; i++)
            g[i].clear(), used[i] = 0;
        for(i = 0; i <= n; i++) {
            for(j = 0; j <= n; j++)
                bg[i][j] = 0xffff; // oo
            bg[i][i] = 0;
        }
        used[n] = 1;
        build(n, n);
        for(k = 0; k <= n; k++)
            for(i = 0; i <= n; i++)
                for(j = 0; j <= n; j++)
                    if(bg[i][j] > bg[i][k]+bg[k][j])
                        bg[i][j] = bg[i][k]+bg[k][j];
        int Q;
        printf("Case %d:\n", ++cases);
        scanf("%d", &Q);
        while(Q--) {
            Pt st, ed;
            scanf("%lf %lf", &st.x, &st.y);
            scanf("%lf %lf", &ed.x, &ed.y);
            int stp = treefind(n, st);
            int edp = treefind(n, ed);
            printf("%d\n", bg[stp][edp]);
        }
    }
    return 0;
}