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  • Chessboard POJ

    there is a pair of integers (x, y) in each line, which represents a hole in the y-th row, the x-th column.

    呃呃呃呃呃。。。因为这个wa了一次

    向周围建边就好了

    #include <iostream>
    #include <cstdio>
    #include <sstream>
    #include <cstring>
    #include <map>
    #include <cctype>
    #include <set>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <bitset>
    #define rap(i, a, n) for(int i=a; i<=n; i++)
    #define rep(i, a, n) for(int i=a; i<n; i++)
    #define lap(i, a, n) for(int i=n; i>=a; i--)
    #define lep(i, a, n) for(int i=n; i>a; i--)
    #define rd(a) scanf("%d", &a)
    #define rlld(a) scanf("%lld", &a)
    #define rc(a) scanf("%c", &a)
    #define rs(a) scanf("%s", a)
    #define rb(a) scanf("%lf", &a)
    #define rf(a) scanf("%f", &a)
    #define pd(a) printf("%d
    ", a)
    #define plld(a) printf("%lld
    ", a)
    #define pc(a) printf("%c
    ", a)
    #define ps(a) printf("%s
    ", a)
    #define MOD 2018
    #define LL long long
    #define ULL unsigned long long
    #define Pair pair<int, int>
    #define mem(a, b) memset(a, b, sizeof(a))
    #define _  ios_base::sync_with_stdio(0),cin.tie(0)
    //freopen("1.txt", "r", stdin);
    using namespace std;
    const int maxn = 1e5 + 10, INF = 0x7fffffff;
    
    int n, m, s, t, k;
    int hol[1100][1100];
    int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    
    int head[maxn], cur[maxn], d[maxn], nex[maxn << 1];
    int cnt;
    
    struct node
    {
        int u, v, c;
    }Node[maxn << 1];
    
    void add_(int u, int v, int c)
    {
        Node[cnt].u = u;
        Node[cnt].v = v;
        Node[cnt].c = c;
        nex[cnt] = head[u];
        head[u] = cnt++;
    }
    
    void add(int u, int v, int c)
    {
        add_(u, v, c);
        add_(v, u, 0);
    }
    
    bool bfs()
    {
        queue<int> Q;
        mem(d, 0);
        Q.push(s);
        d[s] = 1;
        while(!Q.empty())
        {
            int u = Q.front(); Q.pop();
            for(int i = head[u]; i != -1; i = nex[i])
            {
                int v = Node[i].v;
                if(!d[v] && Node[i].c > 0)
                {
                    d[v] = d[u] + 1;
                    Q.push(v);
                    if(v == t) return 1;
                }
            }
        }
        return d[t] != 0;
    }
    
    int dfs(int u, int cap)
    {
        int ret = 0;
        if(u == t || cap == 0) return cap;
        for(int &i = cur[u]; i != -1; i = nex[i])
        {
            int v = Node[i].v;
            if(d[v] == d[u] + 1 && Node[i].c > 0)
            {
                int V = dfs(v, min(Node[i].c, cap));
                Node[i].c -= V;
                Node[i ^ 1].c += V;
                ret += V;
                cap -= V;
                if(cap == 0) break;
            }
        }
        if(cap > 0) d[u] = -1;
        return ret;
    }
    
    int Dinic()
    {
        int ans = 0;
        while(bfs())
        {
            memcpy(cur, head, sizeof head);
            ans += dfs(s, INF);
        }
       return ans;
    }
    
    int main()
    {
        while(scanf("%d%d%d", &n, &m, &k) != EOF)
        {
            mem(head, -1);
            mem(hol, 0);
            cnt = 0;
            s = 0, t = 2000;
            int x, y;
            rap(i, 1, k)
            {
                scanf("%d%d", &y, &x);
                hol[x][y] = 1;
            }
            if((n * m - k) & 1)
            {
                ps("NO");
                continue;
            }
         //   cout << 111 <<endl;
            rap(i, 1, n)
                rap(j, 1, m)
                {
                    if(hol[i][j]) continue;
                    add(s, (i - 1) * m + j, 1);
                    add(1024 + (i - 1) * m + j, t, 1);
                    rep(q, 0, 4)
                    {
                        int nx = i + dir[q][0];
                        int ny = j + dir[q][1];
                        if(nx < 1 || nx > n || ny < 1 || ny > m || hol[nx][ny]) continue;
                        add((i - 1) * m + j, 1024 + (nx - 1) * m + ny, 1);
                    }
                }
           // cout << Dinic() << endl;
          //  cout << n * m
            if(Dinic() == n * m - k)
            {
                ps("YES");
            }
            else
            {
                ps("NO");
            }
    
        }
        return 0;
    }
    自己选择的路,跪着也要走完。朋友们,虽然这个世界日益浮躁起来,只要能够为了当时纯粹的梦想和感动坚持努力下去,不管其它人怎么样,我们也能够保持自己的本色走下去。
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  • 原文地址:https://www.cnblogs.com/WTSRUVF/p/10029253.html
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