URAL

题意：n支队伍打比赛，每2队只进行1场比赛，规定时间内胜得3分，败得0分，若是打到了加时赛，那么胜得2分，败得1分，给出n支队伍最后的总得分，问这个结果是否是可能的，是的话输出“CORRECT”及各场比赛各队伍的比分情况，否则输出"INCORRECT"(2 <= n <= 200)。

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1736

——>>赛后师弟说这是一道网络流大水题，果如其言～

设一个超级源点s，一个超级汇点t，各支队伍各为1个结点，各场比赛也各为1个结点，从s到各场比赛各连1条边，容量为3，从各场比赛到这场比赛的2支参赛队伍各连1条边，容量为3，最后从各支队伍向t各连1条边，容量为输入的对应得分。然后，跑一次最大流，若最大流为满流3 * n * (n-1) / 2，则得分是正确的，再根据各场比赛的流量输出相应的数据，否则得分是不正确的。

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>

using namespace std;

const int maxv = 200 + 10;
const int maxn = 40000 + 10;
const int INF = 0x3f3f3f3f;

int a[maxv], vs[maxv][maxv];

struct Edge{
    int u;
    int v;
    int cap;
    int flow;
};

struct Dinic{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    int addEdge(int uu, int vv, int cap){
        edges.push_back((Edge){uu, vv, cap, 0});
        edges.push_back((Edge){vv, uu, 0, 0});
        m = edges.size();
        G[uu].push_back(m-2);
        G[vv].push_back(m-1);
        return m-2;
    }

    bool bfs(){
        memset(vis, 0, sizeof(vis));
        queue<int> qu;
        qu.push(s);
        d[s] = 0;
        vis[s] = 1;
        while(!qu.empty()){
            int x = qu.front(); qu.pop();
            int si = G[x].size();
            for(int i = 0; i < si; i++){
                Edge& e = edges[G[x][i]];
                if(!vis[e.v] && e.cap > e.flow){
                    vis[e.v] = 1;
                    d[e.v] = d[x] + 1;
                    qu.push(e.v);
                }
            }
        }
        return vis[t];
    }

    int dfs(int x, int a){
        if(x == t || a == 0) return a;
        int flow = 0, f;
        int si = G[x].size();
        for(int& i = cur[x]; i < si; i++){
            Edge& e = edges[G[x][i]];
            if(d[x] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap-e.flow))) > 0){
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if(a == 0) break;
            }
        }
        return flow;
    }

    int Maxflow(int s, int t){
        this->s = s;
        this->t = t;
        int flow = 0;
        while(bfs()){
            memset(cur, 0, sizeof(cur));
            flow += dfs(s, INF);
        }
        return flow;
    }
};

int main()
{
    int n;
    while(scanf("%d", &n) == 1){
        Dinic din;
        int t = n + n * (n-1) / 2 + 1;
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            din.addEdge(i, t, a[i]);
        }
        for(int i = 1, k = n+1; i <= n; i++)
            for(int j = i+1; j <= n; j++, k++){
                vs[i][j] = din.addEdge(0, k, 3);
                din.addEdge(k, i, 3);
                din.addEdge(k, j, 3);
            }
        if(din.Maxflow(0, t) == 3 * n * (n-1) / 2){
            puts("CORRECT");
            for(int i = 1; i <= n; i++)
                for(int j = i+1; j <= n; j++){
                    int L = din.edges[vs[i][j]+2].flow;
                    int R = din.edges[vs[i][j]+4].flow;
                    if(L == 3 && R == 0) printf("%d > %d
", i, j);
                    else if(L == 0 && R == 3) printf("%d < %d
", i, j);
                    else if(L == 2 && R == 1) printf("%d >= %d
", i, j);
                    else printf("%d <= %d
", i, j);
                }
        }
        else puts("INCORRECT");
    }
    return 0;
}