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 C
Weird Fence
Input: Standard Input

Output: Standard Output

In the land of our great Sultan, the World Weird Fence (WWF) festival is going to take place again. For the festival, some poles are set up in a Cartesian plane. Each pole is colored in either red or blue color. You can connect two poles with a chain that consists of multi-colored rings thus creating a weird fence. Each pole has a single hook so you can not connect more than one chain to a pole. Now, though you have an unlimited supply of chains all having the same length, it’s important to note that each of the chains has a red ring at one end & a blue ring at the other end and you are only allowed to hook up a ring to a pole with same color. Also, it’s obvious that you can use a chain to connect two poles if & only if the chain’s length is greater than or equal to the linear distance of those two poles.

Given the co-ordinates of the poles & a positive integer k, write a program to find the minimum possible integer length for the chains so that at least k weird fences can be made. The fences may cross each other.

Input

The first line of the input file is the number of test cases N. This line will be followed by a blank line. N test cases follow. First line of each test case contains two positive integers P & k where P is the number of poles on the plane. (1<=P,k<=100). Each of the next P lines has two integers X & Y & the word “red” / “blue”. X & Y are the co-ordinates of the pole (-1000<=X,Y<=1000) & the word is the color of that pole. No two poles will be in the same location. There will be a blank line between test cases.

Output

For each test case output a single integer in a line which is the minimum integer length of the chains that is necessary to make at least k fences. If it is not possible to build k fences with the given constraints, print the word “Impossible” in a single line.

Problemsetter: Mohammad Mahmudur Rahman

Special thanks To: Manzurur Rahman Khan

題目要求的很特別，直線的歐基里得距離 (Euclidean distance)，卻只要找到 integer 的距離，
使之至少 K 對匹配。

直接拉二分圖，然後二分整數距離，每二分一次就建一次圖進行匹配。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <queue>
using namespace std;
struct Node {
    int x, y, v;// x->y, v
    int next;
} edge[30005];
int e, head[105], dis[105], prev[105], record[105];
void addEdge(int x, int y, int v) {
    edge[e].x = x, edge[e].y = y, edge[e].v = v;
    edge[e].next = head[x], head[x] = e++;
    edge[e].x = y, edge[e].y = x, edge[e].v = 0;
    edge[e].next = head[y], head[y] = e++;
}
int maxflow(int s, int t) {
    int flow = 0;
    int i, j, x, y;
    while(1) {
        memset(dis, 0, sizeof(dis));
        dis[s] =  0xffff; // oo
        queue<int> Q;
        Q.push(s);
        while(!Q.empty()) {
            x = Q.front();
            Q.pop();
            for(i = head[x]; i != -1; i = edge[i].next) {
                y = edge[i].y;
                if(dis[y] == 0 && edge[i].v > 0) {
                    prev[y] = x, record[y] = i;
                    dis[y] = min(dis[x], edge[i].v);
                    Q.push(y);
                }
            }
            if(dis[t])  break;
        }
        if(dis[t] == 0) break;
        flow += dis[t];
        for(x = t; x != s; x = prev[x]) {
            int ri = record[x];
            edge[ri].v -= dis[t];
            edge[ri^1].v += dis[t];
        }
    }
    return flow;
}
struct arc {
    int x, y, v;
    bool operator<(const arc& a) const {
        return v < a.v;
    }
};
int main() {
    int testcase;
    int N, K;
    int i, j, k;
    char s[50];
    arc D[10005];
    int A[10005], Aidx;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d", &N, &K);
        int X[105][2], Y[105][2];
        int nx = 0, ny = 0;
        for(i = 0; i < N; i++) {
            int x, y;
            scanf("%d %d %s", &x, &y, s);
            if(!strcmp(s, "blue")) {
                X[nx][0] = x;
                X[nx][1] = y;
                nx++;
            } else {
                Y[ny][0] = x;
                Y[ny][1] = y;
                ny++;
            }
        }
        int n = 0;
        for(i = 0; i < nx; i++) {
            for(j = 0; j < ny; j++) {
                D[n].x = i+1, D[n].y = nx+j+1;
                D[n].v = (X[i][0]-Y[j][0])*(X[i][0]-Y[j][0])+(X[i][1]-Y[j][1])*(X[i][1]-Y[j][1]);
                int vv = (int)sqrt(D[n].v);
                while(vv*vv < D[n].v)   vv++;
                D[n].v = vv;
                n++;
            }
        }
        sort(D, D+n);
        A[Aidx = 0] = D[0].v;
        for(i = 1; i < n; i++)
            if(D[i].v != A[Aidx])
                A[++Aidx] = D[i].v;
        int l = 0, r = Aidx, mid;
        int ret = 0xfffffff;
        while(l < r) {
            mid = (l+r)/2;
            e = 0;
            memset(head, -1, sizeof(head));
            for(i = 0; i < n && D[i].v <= A[mid]; i++)
                addEdge(D[i].x, D[i].y, 1);
            for(i = 0; i < nx; i++)
                addEdge(0, i+1, 1);
            for(i = 0; i < ny; i++)
                addEdge(nx+i+1, nx+ny+1, 1);
            int flow = maxflow(0, nx+ny+1);
            if(flow >= K)
                r = mid, ret = min(ret, A[mid]);
            else
                l = mid+1;
        }
        if(ret != 0xfffffff)
            printf("%d\n", ret);
        else
            puts("Impossible");
    }
    return 0;
}
/*
2
6 2
-3 5 blue
-3 3 red
1 5 blue
2 0 red
4 6 blue
4 -1 red
*/