• uva12264 Risk


    最小值最大,就二分判断。

    map[i] = '0'+map[i];这样更方便

    每个点拆成ii’,  Sicapa[i]iTcap1(保证至少剩一个)或mid

    i,i’ a[i]  

    i->j’,   inf    //i连到jWA...所以题目的意思大概是只能移动到相邻点?

    判断一下border

    #include<iostream>
    #include<queue>
    #include<cmath>
    #include<cstring>
    #include<vector>
    #include<algorithm>
    #include<string>
    using namespace std;
    const int maxn = 250, INF = 0x3f3f3f3f;
    int a[maxn], border[maxn];
    string map[maxn];
    struct Edge
    {
        int from, to, cap, flow;
        Edge(int u, int v, int c, int f) :from(u), to(v), cap(c), flow(f) {}
    };
    struct EdmondsKarp
    {
        int n, m;
        vector<Edge> edges;
        vector<int> G[maxn];
        int a[maxn];
        int p[maxn];
    
        void init(int n)
        {
            this->n = n;
            for (int i = 0; i < n; i++)
            {
                G[i].clear();
            }
            edges.clear();
            memset(p, 0, sizeof(p));//²»ÐèÒª£¿
        }
        void AddEdge(int u, int v, int c)
        {
            edges.push_back(Edge(u, v, c, 0));
            edges.push_back(Edge(v, u, 0, 0));
            m = edges.size();
            G[u].push_back(m - 2);
            G[v].push_back(m - 1);
        }
        int MaxFlow(int s, int t)
        {
            int flow = 0;
            for (;;)
            {
                memset(a, 0, sizeof(a));
                queue<int> Q;
                Q.push(s);
                a[s] = INF;
                while (!Q.empty())
                {
                    int u = Q.front();
                    Q.pop();
                    for (int i = 0; i < G[u].size(); i++)
                    {
                        Edge& e = edges[G[u][i]];
                        if (!a[e.to] && e.cap > e.flow)
                        {
                            a[e.to] = min(a[u], e.cap - e.flow);
                            Q.push(e.to);
                            p[e.to] = G[u][i];
                        }
                    }
                    if (a[t])
                        break;
                }
                if (!a[t])
                    break;
                for (int i = t; i != s; i = edges[p[i]].from)
                {
                    edges[p[i]].flow += a[t];
                    edges[p[i] ^ 1].flow -= a[t];
                }
                flow += a[t];
            }
            return flow;
        }
    }E;
    
    int Build(int val,int N)
    {
        for (int i = 1; i <= N; i++)
        {
            if (!a[i])
                continue;
            E.AddEdge(0, i, a[i]);
            E.AddEdge(i, i + N, a[i]);
            for (int j = 1; j <= N; j++)
                if (map[i][j] == 'Y')
                {
                    if (!a[j])
                        border[i] = true;         //border
                    else
                        E.AddEdge(i, j + N, INF);
                }
        }
        int ans = 0;
        for (int i = 1; i <= N; i++)
            if (border[i])
            {
                E.AddEdge(i + N, 2 * N + 1, val);
                ans += val;
            }
            else if (a[i])
            {
                E.AddEdge(i + N, 2 * N + 1, 1);
                ans++;
            };
            return ans;
    }
    void solve(int n)
    {
        int a, b, ans;
        int l = 0, r = 10010;
        while (l<r)
        {
            E.init(2 * n + 2);
            int mid = l+(r-l)/2;
            a = Build(mid, n);
            b = E.MaxFlow(0, 2 * n + 1);
            if (a == b)
            {
                l = mid + 1;
                ans = mid;
            }
            else r = mid;
        }
        cout << ans << endl;
    }
    
    void input(int n){
            memset(a, 0, sizeof(a));
            memset(border, 0, sizeof(border));
            for (int i = 0; i < maxn; i++)
                map[i].clear();
    
            for (int i = 1; i <= n; i++)
                cin >> a[i];
            for (int i = 1; i <= n; i++)
            {
                cin >> map[i];
                map[i] = '0'+map[i];
            }
    }
    
    int main()
    {
        int T;
        cin >> T;
        while (T--)
        {
            int n;
            cin >> n;
            input(n);
            solve(n);
        }
        return 0;
    }
  • 相关阅读:
    python复制图片到系统剪切板
    windows 下使用 conda activate 异常
    pandas date_range freq 可选值
    aidlearning aidlux
    python round 四舍五入踩坑
    将JavaBean对象转换为Map集合
    Python输出日志信息
    Feign实现文件上传下载
    常用工具集汇总
    如何对Spring MVC中的Controller进行单元测试
  • 原文地址:https://www.cnblogs.com/lqerio/p/9864914.html
Copyright © 2020-2023  润新知