UVALive

给定一个字符矩阵，s[i][j]为+，-，0分别表示ai+……….aj的和为正，负，0。可以将连续和转化为前缀差，即s[i][j]为+可以表示为B[j]-B[i-1]为正，B[i]表示前i项的和，B[0]为0，然后拓扑排序找出相对关系再依次取值即可。注意相同的B[i]，B[j]要注意保存，最后两者的值统一。最终输出结果a[i] = B[i]-B[i-1]l.

代码：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <string>
#include <map>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#define esp 1e-6
#define pb push_back
#define mp(a, b) make_pair((a), (b))
#define in  freopen("in.txt", "r", stdin);
#define out freopen("out.txt", "w", stdout);
#define print(a) printf("%d
",(a));
#define bug puts("********))))))");
#define stop  system("pause");
#define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
#define inf 0x0f0f0f0f
using namespace std;
typedef long long  LL;
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef vector<pii,int> VII;
typedef vector<int>:: iterator IT;
const int maxn = 20;
VI g[maxn], equa[maxn];
int du[maxn], L[maxn];
int B[maxn];
int n, cnt;
bool toposort(void)
{
    memset(du, 0, sizeof(du));
    for(int i = 0; i <= n; i++)
        for(int j = 0; j < g[i].size(); j++)
            du[g[i][j]]++;
    queue<int>  Q;
    for(int i = 0; i <= n; i++)
        if(!du[i])
            Q.push(i);
    cnt = 0;
    while(!Q.empty())
    {
        int x = Q.front();
        Q.pop();
        L[cnt++] = x;
        for(int i = 0; i < g[x].size(); i++)
        {
            int v = g[x][i];
            du[v]--;
            if(!du[v])
                Q.push(v);
        }
    }
    return cnt == n+1;
}
int main(void)
{
    int T;
    for(int t = scanf("%d", &T); t <= T; t++)
    {
        for(int i = 0; i < maxn; i++)
            g[i].clear(),equa[i].clear();
        memset(B, 0x0f, sizeof(B));
        scanf("%d", &n);
        char s[1000];
        scanf("%s", s);
        cnt = 0;
        for(int i = 1; i <= n; i++)
            for(int j = i; j <= n; j++)
            {
                if(s[cnt] == '+')  g[i-1].pb(j);
                else if(s[cnt] == '-')  g[j].pb(i-1);
                else   equa[i-1].pb(j);
                cnt++;
            }
        toposort();
        B[0] = 0;
        int k;
        for(int i = 0; i < cnt; i++)
            if(!B[L[k = i]]) break;
        for(int i = k -1; i >= 0; i--)
            B[L[i]] = i - k;
        for(int i = k+1; i < cnt; i++)
            B[L[i]] = i-k;
        for(int i = 0; i < cnt; i++)
            for(int j = 0; j < equa[L[i]].size(); j++)
                B[equa[L[i]][j]] = B[L[i]];
        for(int i = 1; i <= n; i++)
            printf("%d ", B[i]-B[i-1]);
        puts("");
    }
    return 0;
}