UVALive 4255 前缀和 拓扑排序

题意：定义为的正负号，给出符号矩阵，需要求出对应的一个序列。<=10，序列的长度n<=10
解法：首先根据连续和转化为前缀和之差的思想，设，则-，那么矩阵中的一项表示了两个之间的大小关系，设=0，问题转化为已知之间的大小关系，求出一个对应的序列。用拓扑排序求出它们的关系，构造一组解，然后很容易得出序列

#include
using namespace std;

const int maxn = 15;
int f[maxn][maxn];
int topo[maxn];
int vis[maxn],temp;
int n;
int b[maxn];

bool dfs(int u){
    vis[u] = -1;
    for(int i = 0; i <= n; ++i){
        if(f[u][i] == 1){
            if(vis[i] < 0) return false;
            if(!vis[i] && !dfs(i)) return false;
        }
    }
    vis[u] = 1; topo[--temp] = u;
    return true;
}

bool toposort(){
    temp = n+1;
    for(int i = 0; i <= n; ++i){
        if(!vis[i]){
            if(!dfs(i)){
                return false;
            }
        }
    }
    return true;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        memset(vis,0,sizeof(vis));
        memset(f,0,sizeof(f));
        memset(topo,0,sizeof(topo));
        memset(b,0,sizeof(b));
        scanf("%d",&n);
        char c;
        getchar();
        for(int i = 1; i <= n; ++i){
            for(int j = i; j <= n; ++j){
                scanf("%c",&c);
                if(c=='+') f[i-1][j] = 1;
                else if(c=='-') f[j][i-1] = 1;
                else if(c=='0') f[i-1][j] = f[j][i-1] = -1;
            }
        }
        toposort();
        int loc0;
        for(int i = 0; i <= n; ++i){
            if(topo[i] == 0){
                loc0 = i;
                break;
            }
        }
        int now = 0;
        for(int i = loc0+1; i <= n; ++i){
            if(f[topo[i]][topo[i-1]] != -1) b[topo[i]] = ++now;
            else b[topo[i]] = now;
        }
        now = 0;
        for(int i = loc0-1; i >= 0; --i){
            if(f[topo[i]][topo[i+1]] != -1) b[topo[i]] = --now;
            else b[topo[i]] = now;
        }
        for(int i = 1; i <= n; ++i){
            printf("%d%c",b[i]-b[i-1]," \n"[i==n?1:0]);
        }
    }
    return 0;
}