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  • K

    题目链接:https://vjudge.net/contest/299467#problem/K

     

    这个题目从数据范围来看可以发现是网络流,怎么建图呢?这个其实不是特别难,主要是读题难。

    这个建图就是把源点和每一个蜥蜴存在的点相连,汇点和可以跑出去的相连,因为这个题目对于每一个点都有次数要求,所以就要拆点。

     

     

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <string>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #define inf 0x3f3f3f3f
    using namespace std;
    typedef long long ll;
    const int INF = 0x3f3f3f3f;
    const int maxn = 5e5 + 10;
    struct edge
    {
        int u, v, c, f;
        edge(int u, int v, int c, int f) :u(u), v(v), c(c), f(f) {}
    };
    vector<edge>e;
    vector<int>G[maxn];
    int level[maxn];//BFS分层,表示每个点的层数
    int iter[maxn];//当前弧优化
    int m;
    void init()
    {
        for (int i = 0; i <= maxn; i++)G[i].clear();
        e.clear();
    }
    void add(int u, int v, int c)
    {
        e.push_back(edge(u, v, c, 0));
        e.push_back(edge(v, u, 0, 0));
        m = e.size();
        G[u].push_back(m - 2);
        G[v].push_back(m - 1);
    }
    void BFS(int s)//预处理出level数组
    //直接BFS到每个点
    {
        memset(level, -1, sizeof(level));
        queue<int>q;
        level[s] = 0;
        q.push(s);
        while (!q.empty())
        {
            int u = q.front();
            q.pop();
            for (int v = 0; v < G[u].size(); v++)
            {
                edge& now = e[G[u][v]];
                if (now.c > now.f && level[now.v] < 0)
                {
                    level[now.v] = level[u] + 1;
                    q.push(now.v);
                }
            }
        }
    }
    int dfs(int u, int t, int f)//DFS寻找增广路
    {
        if (u == t)return f;//已经到达源点,返回流量f
        for (int &v = iter[u]; v < G[u].size(); v++)
            //这里用iter数组表示每个点目前的弧,这是为了防止在一次寻找增广路的时候,对一些边多次遍历
            //在每次找增广路的时候,数组要清空
        {
            edge &now = e[G[u][v]];
            if (now.c - now.f > 0 && level[u] < level[now.v])
                //now.c - now.f > 0表示这条路还未满
                //level[u] < level[now.v]表示这条路是最短路,一定到达下一层,这就是Dinic算法的思想
            {
                int d = dfs(now.v, t, min(f, now.c - now.f));
                if (d > 0)
                {
                    now.f += d;//正向边流量加d
                    e[G[u][v] ^ 1].f -= d;
                    //反向边减d,此处在存储边的时候两条反向边可以通过^操作直接找到
                    return d;
                }
            }
        }
        return 0;
    }
    int Maxflow(int s, int t)
    {
        int flow = 0;
        for (;;)
        {
            BFS(s);
            if (level[t] < 0)return flow;//残余网络中到达不了t,增广路不存在
            memset(iter, 0, sizeof(iter));//清空当前弧数组
            int f;//记录增广路的可增加的流量
            while ((f = dfs(s, t, INF)) > 0)
            {
                flow += f;
            }
        }
        return flow;
    }
    
    struct node
    {
        int x, y, flow;
    }exa[maxn];
    
    int main()
    {
        int qaq, cas = 0;
        scanf("%d", &qaq);
        while(qaq--)
        {
            init();
            int nn, dd, tot = 0, len;
            char cs[110];
            scanf("%d%d", &nn, &dd);
            for(int i=0;i<nn;i++)
            {
                scanf("%s", cs);
                len = strlen(cs);
                for(int j=0;j<len;j++)
                {
                    exa[++tot].x = i;
                    exa[tot].y = j;
                    exa[tot].flow = cs[j] - '0';
                }
            }
            char mp[110][110];
            int s = 0, t = tot * 2 + 1;
            for(int i=0;i<nn;i++)
            {
                scanf("%s", mp[i]);
            }
            int sum = 0;
            for(int i=1;i<=tot;i++)
            {
                if(exa[i].flow>0)
                {
                    add(i, i + tot, exa[i].flow);
    
                    if (mp[exa[i].x][exa[i].y] == 'L')
                    {
                        sum++;
                        add(s, i, 1);
                    }
                    if (exa[i].x < dd || exa[i].y < dd || (nn - exa[i].x) <= dd || (len - exa[i].y) <= dd) add(i + tot, t, inf);
    
                    for(int j=1;j<=tot;j++)
                    {
                        if (i == j) continue;
                        int dis = (exa[i].x - exa[j].x)*(exa[i].x - exa[j].x) + (exa[i].y - exa[j].y)*(exa[i].y - exa[j].y);
                        if (exa[j].flow&&dis <= dd*dd)
                        {
                            add(i+tot, j, inf);
                        }
                    }
                }
            }
            int ans = Maxflow(s, t);
            sum = sum - ans;
            if (sum == 0) printf("Case #%d: no lizard was left behind.
    ",++cas);
            else if (sum == 1) printf("Case #%d: 1 lizard was left behind.
    ",++cas);
            else printf("Case #%d: %d lizards were left behind.
    ",++cas,sum);
        }
        return 0;
    }

     

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <string>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #define inf 0x3f3f3f3f
    using namespace std;
    typedef long long ll;
    const int INF = 0x3f3f3f3f;
    const int maxn = 5e5 + 10;
    struct edge
    {
        int u, v, c, f;
        edge(int u, int v, int c, int f) :u(u), v(v), c(c), f(f) {}
    };
    vector<edge>e;
    vector<int>G[maxn];
    int level[maxn];//BFS分层,表示每个点的层数
    int iter[maxn];//当前弧优化
    int m;
    void init()
    {
        for (int i = 0; i <= maxn; i++)G[i].clear();
        e.clear();
    }
    void add(int u, int v, int c)
    {
        e.push_back(edge(u, v, c, 0));
        e.push_back(edge(v, u, 0, 0));
        m = e.size();
        G[u].push_back(m - 2);
        G[v].push_back(m - 1);
    }
    void BFS(int s)//预处理出level数组
    //直接BFS到每个点
    {
        memset(level, -1, sizeof(level));
        queue<int>q;
        level[s] = 0;
        q.push(s);
        while (!q.empty())
        {
            int u = q.front();
            q.pop();
            for (int v = 0; v < G[u].size(); v++)
            {
                edge& now = e[G[u][v]];
                if (now.c > now.f && level[now.v] < 0)
                {
                    level[now.v] = level[u] + 1;
                    q.push(now.v);
                }
            }
        }
    }
    int dfs(int u, int t, int f)//DFS寻找增广路
    {
        if (u == t)return f;//已经到达源点,返回流量f
        for (int &v = iter[u]; v < G[u].size(); v++)
            //这里用iter数组表示每个点目前的弧,这是为了防止在一次寻找增广路的时候,对一些边多次遍历
            //在每次找增广路的时候,数组要清空
        {
            edge &now = e[G[u][v]];
            if (now.c - now.f > 0 && level[u] < level[now.v])
                //now.c - now.f > 0表示这条路还未满
                //level[u] < level[now.v]表示这条路是最短路,一定到达下一层,这就是Dinic算法的思想
            {
                int d = dfs(now.v, t, min(f, now.c - now.f));
                if (d > 0)
                {
                    now.f += d;//正向边流量加d
                    e[G[u][v] ^ 1].f -= d;
                    //反向边减d,此处在存储边的时候两条反向边可以通过^操作直接找到
                    return d;
                }
            }
        }
        return 0;
    }
    int Maxflow(int s, int t)
    {
        int flow = 0;
        for (;;)
        {
            BFS(s);
            if (level[t] < 0)return flow;//残余网络中到达不了t,增广路不存在
            memset(iter, 0, sizeof(iter));//清空当前弧数组
            int f;//记录增广路的可增加的流量
            while ((f = dfs(s, t, INF)) > 0)
            {
                flow += f;
            }
        }
        return flow;
    }
    
    struct node
    {
        int x, y, flow;
    }exa[maxn];
    
    int main()
    {
        int qaq, cas = 0;
        scanf("%d", &qaq);
        while (qaq--)
        {
            init();
            int nn, dd, tot = 0, len;
            char cs[110];
            scanf("%d%d", &nn, &dd);
            for (int i = 1; i <= nn; i++)
            {
                scanf("%s", cs+1);
                len = strlen(cs+1);//这个地方要注意一下,就是因为这个wa了几发,这个地方不可以写strlen(cs)-1
                for (int j = 1; j <= len; j++)
                {
                    exa[++tot].x = i;
                    exa[tot].y = j;
                    exa[tot].flow = cs[j] - '0';
                }
            }
            char mp[110][110];
            int s = 0, t = tot * 2 + 1;
            for (int i = 1; i <= nn; i++)
            {
                scanf("%s", mp[i]+1);
            }
            int sum = 0;
            for (int i = 1; i <= tot; i++)
            {
                if (exa[i].flow > 0)
                {
                    add(i, i + tot, exa[i].flow);
    
                    if (mp[exa[i].x][exa[i].y] == 'L')
                    {
                        sum++;
                        add(s, i, 1);
                    }
                    if (exa[i].x <= dd || exa[i].y <= dd || (nn - exa[i].x) < dd || (len - exa[i].y) < dd) add(i + tot, t, inf);
    
                    for (int j = 1; j <= tot; j++)
                    {
                        if (i == j) continue;
                        int dis = (exa[i].x - exa[j].x)*(exa[i].x - exa[j].x) + (exa[i].y - exa[j].y)*(exa[i].y - exa[j].y);
                        if (exa[j].flow&&dis <= dd * dd)
                        {
                            add(i + tot, j, inf);
                        }
                    }
                }
            }
            int ans = Maxflow(s, t);
            sum = sum - ans;
            if (sum == 0) printf("Case #%d: no lizard was left behind.
    ", ++cas);
            else if (sum == 1) printf("Case #%d: 1 lizard was left behind.
    ", ++cas);
            else printf("Case #%d: %d lizards were left behind.
    ", ++cas, sum);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/EchoZQN/p/10830961.html
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