UValive 4255 Guess（拓扑排序）

TAT~图论好难，3个月前就见过这道题了，当初拿过来这题根本没感觉是个图论。

而后听了嘉神简单讲了下思路也是糊里糊涂，只知道是个拓扑排序问题。如今再看到这题依旧没个敲的思路。

直到盯着队友的代码看了半个小时，又手跑了一趟，才终于明白个所以然，感觉好神奇= =|||

【虽然能感觉到自己有进步，但真不知道什么时候才能把自己这颗“雪球”滚的越来越大】

Description

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:

 12341-+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

解题思路：首先是要把求每个位置上的值转化为求 “前缀和之差”。

input数组中，input[i]中存的值，表示有多少个"前缀和"比"前i个数的和"大。

vec数组中，vec[i]中存的值x，表示"前x数的和"比"前i个小"。

然后假设最大的前缀和为20，依次第二大19,第三大18递减得到。

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <stack>#include <string>#include <iostream>using namespace std;#define maxn 15char ch[maxn * maxn];int t, n, k, ans;int input[maxn], sum[maxn];vector<int> vec[maxn];void init() {    k = 0; ans = 20;    memset(input, 0, sizeof(input));    memset(sum, 0, sizeof(sum));    for(int i = 0; i <= n; i++)        vec[i].clear();}void add_edge() {    for(int i = 1; i <= n; i++) {        for(int j = i; j <= n; j++) {            char temp = ch[k++];            if(temp == '+') {                input[i - 1]++;                vec[j].push_back(i - 1);            }            else if(temp == '-') {                input[j]++;                vec[i - 1].push_back(j);            }        }    }}void topsort() {    stack<int> s;    for(int i = 0; i <= n; i++)        if(input[i] == 0) {            s.push(i);            sum[i] = ans;            input[i] = -1;        }    while(!s.empty()) {        int x = s.top(); s.pop();        ans--; input[x] = -1;        for(int i = 0; i < vec[x].size(); i++) {            input[vec[x][i]]--;            if(input[vec[x][i]] == 0) {                s.push(vec[x][i]);                input[vec[x][i]] = -1;                sum[vec[x][i]] = ans;            }        }    }}int main() {    scanf("%d", &t);    while(t--) {        scanf("%d%s", &n, ch);        init();        add_edge();        topsort();        for(int i = 1; i <= n; i++) {            if(i == 1) printf("%d", sum[i] - sum[i - 1]);            else printf(" %d", sum[i] - sum[i - 1]);        }        printf("\n");    }    return 0;}