算法——图论——关键活动

原题

cpp 复制代码
#include <iostream>
#include <vector>
#include <queue>
#include <set>

using namespace std;

struct edge {
    int destination;
    int dist;

    edge(int destination_, int dist_) : destination(destination_), dist(dist_) {}
};

vector<edge> graph[100];
vector<edge> reGraph[100];
vector<int> inDo(100, 0);
vector<int> outDo(100, 0);
vector<int> lessTime(100, 0);
vector<int> moreTime(100, 0x3fffffff);

int main() {

    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        int u, v, w;
        cin >> u >> v >> w;
        graph[u].emplace_back(v, w);
        reGraph[v].emplace_back(u, w);
        outDo[u]++;
        inDo[v]++;
    }
    queue<int> q;
    for (int i = 0; i < n; ++i) {
        if (inDo[i] == 0) {
            q.push(i);
            lessTime[i] = 0;
        }
    }

    while (!q.empty()) {
        int cur = q.front();
        q.pop();

        for (edge neighbor: graph[cur]) {
            inDo[neighbor.destination]--;
            lessTime[neighbor.destination] = max(lessTime[neighbor.destination], lessTime[cur] + neighbor.dist);
            if (inDo[neighbor.destination] == 0) {
                q.push(neighbor.destination);
            }
        }
    }

    int totalTime = 0;
    for (int i = 0; i < n; ++i) {
        if (inDo[i] != 0) {
            cout << "No" << endl;
            return 0;
        }
        totalTime = max(totalTime, lessTime[i]);
    }

    for (int i = 0; i < n; ++i) {
        if (outDo[i] == 0) {
            q.push(i);
            moreTime[i] = totalTime;
        }
    }


    while (!q.empty()) {
        int cur = q.front();
        q.pop();

        for (edge neighbor: reGraph[cur]) {
            outDo[neighbor.destination]--;
            moreTime[neighbor.destination] = min(moreTime[neighbor.destination], moreTime[cur] - neighbor.dist);
            if (outDo[neighbor.destination] == 0) {
                q.push(neighbor.destination);
            }
        }
    }

    set<pair<int, int>> s;
    for (int i = 0; i < n; ++i) {
    for (edge neighbor: graph[i]) {
        if (moreTime[neighbor.destination] - lessTime[i] - neighbor.dist == 0) {
            s.insert({i, neighbor.destination});
        }
    }
}

//    for (int i = 0; i < n; ++i) {
//        if (lessTime[i] == 0 && moreTime[i] == 0) {
//            q.push(i);
//        }
//    }

//    while (!q.empty()) {
//        auto cur = q.front();
//        q.pop();

//        for (auto neighbor: graph[cur]) {
//            if (lessTime[neighbor.destination] == moreTime[neighbor.destination]) {
//                s.insert({cur, neighbor.destination});
//            q.push(neighbor.destination);
//            }
//        }
//    }

    cout << "Yes" << endl;
    for (auto p: s) {
        cout << p.first << " " << p.second << endl;
    }
    return 0;
}
相关推荐
爱刷碗的苏泓舒3 小时前
PPP-AR 中的参考星选取:数学原理、评价指标与切换处理
算法·gnss·模糊度固定·ppp-ar·星间单差·参考星·卫星端偏差
烬羽6 小时前
递归老写崩?一个"退回"公式,把回溯题变成填空题
javascript·深度学习·算法
闪电悠米6 小时前
力扣hot100-41.缺失的第一个正数-原地哈希详解
数据结构·算法·哈希算法
星空露珠6 小时前
28种颜色对应名称,
开发语言·数据库·算法·游戏·lua
QXWZ_IA8 小时前
桥梁数字孪生怎么落地?
人工智能·科技·算法·智能硬件·政务
Kel8 小时前
输出层与反分词(Output Layer & Detokenization)
人工智能·算法·架构
陕西企来客8 小时前
2026年7月技术好GEO优化方案:算法适配与内容策略
人工智能·算法·机器学习·技术好geo优化
风栖柳白杨9 小时前
【面试】AI算法工程师_空白自测版本
人工智能·算法·面试
闪电悠米10 小时前
力扣hot100-73.矩阵置零-标记数组详解
算法·leetcode·矩阵
栋***t10 小时前
从“纸质试卷”到“AI智能组卷”,麦塔在线考试系统如何重构出题逻辑?
java·大数据·人工智能·算法·重构