ZOJ 2856 Happy Life 暴力求解

因为是Special Judge 的题目，只要输出正确答案即可，不唯一

暴力力求解, 只要每次改变 happiness 值为负的人的符号即可。

如果计算出当前人的 happiness 值为负，那么将其 p(i) 值赋值为-p(i) 即可
这题是保证有解的,至至于为何难以证明。

 

Source Code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <set>

#include <list>

#include <queue>

#include <vector>

#include <algorithm>

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define pi acos(-1.0)



using namespace std;



typedef long long           ll      ;

typedef unsigned long long  ull     ;

typedef unsigned int        uint    ;

typedef unsigned char       uchar   ;



template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}

template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}



const double eps = 1e-7      ;

const int N = 210            ;

const int M = 1100011*2      ;

const ll P = 10000000097ll   ;

const int MAXN = 10900000    ;



int a[220][220], n;

int ans[220];



int main(){

    std::ios::sync_with_stdio(false);

    int i, j, t, k, u, v, numCase = 0;



    while (EOF != scanf ("%d",&n)) {

        for (i = 1; i <= n; ++i) {

            for (j = 1; j <= n; ++j) {

                scanf("%d", &a[i][j]);

            }

        }

        for (i = 1; i <= n; ++i)    ans[i] = 1;

        int cnt = 1;

        for (;;) {

            if (cnt > n)   break;

            int sum = 0;

            for (i = 1; i <= n; ++i) {

                sum += a[cnt][i] * ans[i];

            }

            if (sum * ans[cnt] < 0) {

                ans[cnt] = -ans[cnt];

                cnt = 1;

            } else {

                ++cnt;

            }

        }

        printf ("Yes\n");

        for (i = 1; i <= n; ++i) {

            if (ans[i] == 1) {

                printf("+\n");

            } else {

                printf("-\n");

            }

        }

    }



    return 0;

}