P4819 [中山市选]杀人游戏

题目描述

一位冷血的杀手潜入Na-wiat，并假装成平民。警察希望能在NN个人里面，查出谁是杀手。警察能够对每一个人进行查证，假如查证的对象是平民，他会告诉警察，他认识的人，谁是杀手，谁是平民。假如查证的对象是杀手，杀手将会把警察干掉。现在警察掌握了每一个人认识谁。每一个人都有可能是杀手，可看作他们是杀手的概率是相同的。

问：根据最优的情况，保证警察自身安全并知道谁是杀手的概率最大是多少？

Solution

首先缩点, 然后需要把缩完点后的DAG上每个入度为0的点都询问一次才行.
但是有一种特殊情况是有一个入度为0的点, 它连接的点都不是必须需要它.这时就可以不询问它了.

Code

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const int N = 1e5 + 5, M = 3e5 + 6;
class BaseGraph {
public:
    struct Edge {
        int v; Edge* nxt;
        Edge(int _, Edge* __) : v(_), nxt(__) { }
    } *head[N];
    int du[N];
    BaseGraph() {
        memset(du, false, sizeof du);
        for (int i = 1; i < N; i += 1)
            head[i] = nullptr;
    }
    void AddEdge(int u, int v) {
        du[v] += 1; head[u] = new Edge(v, head[u]);
    }
};
class Graph : public BaseGraph {
    int dfn[N], low[N], vis[N], que[N], col[N], siz[N];
    int vis_num, col_num, top;
    void Tarjan(int u) {
        dfn[u] = low[u] = ++vis_num;
        vis[u] = true, que[++top] = u;
        for (auto edge = head[u]; edge; edge = edge->nxt) {
            if (not dfn[edge->v])
                Tarjan(edge->v), low[u] = std:: min(low[u], low[edge->v]);
            else if (vis[edge->v])
                low[u] = std:: min(low[u], low[edge->v]);
        }
        if (dfn[u] == low[u]) {
            vis[u] = false, col[u] = ++col_num, siz[col_num] = 1;
            while (que[top] != u)
                vis[que[top]] = false,
                col[que[top--]] = col_num, siz[col_num] += 1;
            top--;
        }
    }
public:
    double init(int n, int m) {
        for (int i = 0, u, v; i < m; i += 1) {
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
        }
        for (int i = 1; i <= n; i += 1)
            if (not dfn[i])
                Tarjan(i);
        BaseGraph* rG = new BaseGraph();
        for (int u = 1; u <= n; u += 1)
            for (auto edge = head[u]; edge; edge = edge->nxt)
                if (col[u] != col[edge->v])
                    rG->AddEdge(col[u], col[edge->v]);
        int num_without_du = 0;
        for (int i = 1; i <= col_num; i += 1)
            if (not rG->du[i])
                num_without_du += 1;
        for (int u = 1; u <= n; u += 1)
            if (not rG->du[col[u]] and siz[col[u]] == 1) {
                bool flag = false;
                for (auto edge = head[u]; edge; edge = edge->nxt) {
                    if (rG->du[col[edge->v]] == 1) {
                        flag = true; break;
                    }
                }
                if (not flag) {
                    return 1.0 - 1.0 * (num_without_du - 1) / n;
                }
            }
        return 1.0 - 1.0 * num_without_du / n;
    }

};

int main () {
    int n, m;
    Graph* G = new Graph();
    scanf("%d%d", &n, &m);
    printf("%.6f\n", G->init(n, m));
    return 0;
}

原文地址：https://www.cnblogs.com/qdscwyy/p/9835474.html