uva 1423 / Guess

Given a sequence of integers, a₁, a₂,..., a_n , we define its sign matrix S such that, for 1ijn , S_ij = `` + " if a<sub>i</sub> +...+ a<sub>j</sub> > 0 ; S<sub>ij</sub> = `` - " if a_i +...+ a_j < 0 ; and S_ij = ``0" otherwise.

For example, if (a₁, a₂, a₃, a₄) = (- 1, 5, - 4, 2) , then its sign matrix S is a 4×4 matrix:

 

	1	2	3	4
1	-	+	0	+
2		+	+	+
3			-	-
4				+

 

We say that the sequence (-1, 5, -4, 2) generates the sign matrix. A sign matrix is valid if it can be generated by a sequence of integers.

Given a sequence of integers, it is easy to compute its sign matrix. This problem is about the opposite direction: Given a valid sign matrix, find a sequence of integers that generates the sign matrix. Note that two or more different sequences of integers can generate the same sign matrix. For example, the sequence (-2, 5, -3, 1) generates the same sign matrix as the sequence (-1,5, -4,2).

Write a program that, given a valid sign matrix, can find a sequence of integers that generates the sign matrix. You may assume that every integer in a sequence is between -10 and 10, both inclusive.

 

Input 

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case consists of two lines. The first line contains an integer n (1n10) , where n is the length of a sequence of integers. The second line contains a string of n(n + 1)/2 characters such that the first n characters correspond to the first row of the sign matrix, the next n - 1 characters to the second row, ... , and the last character to the n -th row.

 

Output 

Your program is to write to standard output. For each test case, output exactly one line containing a sequence of n integers which generates the sign matrix. If more than one sequence generates the sign matrix, you may output any one of them. Every integer in the sequence must be between -10 and 10, both inclusive.

 

Sample Input 

 

3 
4 
-+0++++--+ 
2 
+++ 
5 
++0+-+-+--+-+--

 

Sample Output 

 

-2 5 -3 1 
3 4 
1 2 -3 4 -5

LRJ的代码看不懂。。。然后发现http://blog.csdn.net/wiking__acm/article/details/8660931这位大神的代码和解说都比较易懂，于是照敲了一遍，很纠结地A了。
其实就是拓扑排序，删除入度为零的点，删除其指出的边，开始觉得有点像差分约束题，但差分约束貌似只能判断有无解，而无法构造解。。。不过问题建模是一样的，不等式组，构造连接图：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<cmath>
#include<fstream>
#include<iomanip>
#include<climits>
#include<cfloat>

using namespace std;

#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define ULL unsigned long long
#define LL long long
#define MAX(x,y) ((x) > (y) ? (x) : (y))
#define MIN(x,y) ((x) < (y) ? (x) : (y))

#define MAXN 13
int n;
int map[MAXN][MAXN];
int degree[MAXN];
int s[MAXN];
char text[1000];

void Topsort(){
    int cnt=0,cur=-10;
    while(cnt<n+1){
        int flag[MAXN];
        memset(flag, 0, sizeof(flag));          //标记出入度为0的点
        for(int i=0; i<=n; i++) if(!degree[i]){
            degree[i]=-1;
            flag[i]=1;
            s[i]=cur;                           //构造解
            cnt++;
        }
        cur++;                                  //删除由标记点指出的弧
        for(int i=0; i<=n; i++) if(flag[i]){    
            for(int j=0; j<=n; j++) if(map[i][j]){
                degree[j]--;
            }
        }
    }
}

void ini(){
    memset(map, 0, sizeof(map));
    memset(degree, 0, sizeof(degree));
}

int main(){
    //freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    int cas;
    scanf(" %d",&cas);
    while(cas--){
        scanf(" %d%s",&n,text);
        ini();
        int c=0;
        for(int i=1; i<=n; i++)
            for(int j=i; j<=n; j++){
                if(text[c]=='+'){               //加边，s[j]-s[i-1]>0则i-1指向j,
                    map[i-1][j]=1;  degree[j]++;
                }
                else if(text[c]=='-'){
                    map[j][i-1]=1;  degree[i-1]++;
                }
                c++;
            }
        Topsort();                              //拓扑排序
        for(int i=1; i<=n; i++)
            printf("%d%c",s[i]-s[i-1],(i==n ? '\n':' '));
    }
    return 0;
}