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 G

Golden Tiger Claw

Time Limit: 8 Second

Omi, Raymondo, Clay and Kimiko are on new adventure- in search of new Shen Gong Wu. But Evil Boy Genius Jack Spicer is also there. Omi and Jack found the Shen Gong Wu at the same time so they rushed for it but alas they touched it at the same time. Then what? It is time for “Xiaolin Showdown”.

Jack challenged Omi to play a game. The game is simple! There will be an N*N board where each cell in the board contains some number. They have to assign numbers to each row and column separately so that  where  is the number assigned to the cell located at i’th row and j’th column,  is the number assigned to i’th row and  is the number assigned to j’th column. That is simple isn’t it? Well… the main part is that you have to minimize .

Jack has taken his favorite “Monkey Stuff” and Omi has taken “Golden Tiger Claw”. With the help of this “Golden Tiger Claw”, he can go anywhere in the world. He has come to you and seeking your help. Jack is using his computer to solve this problem. So do it quick! Find the most optimal solution for Omi so that you can also be part of history in saving the world from the darkness of evil.

Input:

Input contains 15 test cases. Each case starts with N. Then there are N lines containing N numbers each. All the numbers in input is positive integer within the limit 100 except N which can be at most 500.

Output:

For each case in the first line there will be N numbers, the row assignments. In the next line there will N column assignment. And at the last line the minimum sum should be given. If there are several possible solutions give any.

Problemsetter: Md. Mahbubul Hasan

(Be careful about the output format. You may get Wrong Answer if you don’t output properly)

KM 算法的核心操作，而這一道問題剛好是副產物。

詳見 : http://www.cnblogs.com/silver-bullet/archive/2013/02/22/2922064.html

#include <stdio.h>
#include <string.h>
int W[505][505], N;
int mx[505], my[505]; // match arr
int lx[505], ly[505]; // label arr
int x[505], y[505]; // used arr
int hungary(int nd) {
    int i;
    x[nd] = 1;
    for(i = 1; i <= N; i++) {
        if(y[i] == 0 && W[nd][i] == lx[nd]+ly[i]) {
            y[i] = 1;
            if(my[i] == 0 || hungary(my[i])) {
                my[i] = nd;
                return 1;
            }
        }
    }
    return 0;
}
int KM() {
    int i, j, k, d;
    memset(mx, 0, sizeof(mx));
    memset(my, 0, sizeof(my));
    memset(lx, 0, sizeof(lx));
    memset(ly, 0, sizeof(ly));
    for(i = 1; i <= N; i++)
        for(j = 1; j <= N; j++)
            lx[i] = lx[i] > W[i][j] ? lx[i] : W[i][j];
    for(i = 1; i <= N; i++) {
        while(1) {
            memset(x, 0, sizeof(x));
            memset(y, 0, sizeof(y));
            if(hungary(i))  break;
            d = 0xfffffff;
            for(j = 1; j <= N; j++) {
                if(x[j]) {
                    for(k = 1; k <= N; k++)
                        if(!y[k])
                        d = d < lx[j]+ly[k]-W[j][k] ?
                            d : lx[j]+ly[k]-W[j][k];
                }
            }
            if(d == 0xfffffff)  break;
            for(j = 1; j <= N; j++) {
                if(x[j])    lx[j] -= d;
                if(y[j])    ly[j] += d;
            }
        }
    }
    int res = 0;
    for(i = 1; i <= N; i++) {
        if(my[i])
            res += W[my[i]][i];
    }
    return res < 0 ? 0 : res;
}
int main() {
    int n;
    while(scanf("%d", &n) == 1 && n) {
        N = n;
        int i, j;
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
                scanf("%d", &W[i][j]);
        KM();
        int ret = 0;
        for(i = 1; i <= n; i++)
            printf("%d%c", lx[i], i == n ? '\n' : ' '), ret += lx[i];
        for(i = 1; i <= n; i++)
            printf("%d%c", ly[i], i == n ? '\n' : ' '), ret += ly[i];
        printf("%d\n", ret);
       
    }
    return 0;
}