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\n", i, j);else if(L == 0 && R == 3) printf("%d < %d\n", i, j);else if(L == 2 && R == 1) printf("%d >= %d\n", i, j);else printf("%d <= %d\n", i, j);}}else puts("INCORRECT");}return 0;
}