URAL - 1736 - Chinese Hockey

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;

}