P4878 [USACO05DEC] Layout G

合集 - 图论(79)

1.F. Chat Screenshots2024-02-192.P1656 炸铁路2024-02-223.P1137 旅行计划2024-02-244.P2835 刻录光盘2024-02-245.P1197 [JSOI2008] 星球大战2024-02-246.P3388 【模板】割点（割顶）2024-02-247.P8435 【模板】点双连通分量2024-02-258.P8436 【模板】边双连通分量2024-02-259.P2860 [USACO06JAN] Redundant Paths G2024-02-2510.P1653 [USACO04DEC] Cow Ski Area G2024-02-2611.P3047 [USACO12FEB] Nearby Cows G2024-03-0412.P1894 [USACO4.2] 完美的牛栏The Perfect Stall2024-03-0613.P1550 [USACO08OCT] Watering Hole G2024-03-0614.P2330 [SCOI2005] 繁忙的都市2024-03-0615.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0716.P1379 八数码难题2024-03-0717.P2746 [USACO5.3] 校园网Network of Schools2024-03-0818.P6121 [USACO16OPEN] Closing the Farm G2024-03-0819.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0820.P2055 [ZJOI2009] 假期的宿舍2024-03-0821.P5905 【模板】全源最短路（Johnson）2024-03-1122.F. Microcycle2024-03-1223.G. Path Prefixes2024-03-1424.G. Rudolf and Subway2024-03-1525.C. Ehab and Path-etic MEXs2024-03-1726.A. String Transformation 12024-03-1727.D. Secret Passwords2024-03-1928.F. Maximum White Subtree2024-03-1929.P3478 [POI2008] STA-Station2024-03-1930.P1347 排序2024-03-2131.P1960 郁闷的记者2024-03-2132.E1. Weights Division (easy version)2024-03-2233.P5007 DDOSvoid 的疑惑2024-03-2534.P2850 [USACO06DEC] Wormholes G2024-03-2635.P1265 公路修建2024-03-2836.P1354 房间最短路问题2024-03-2937.P2168 [NOI2015] 荷马史诗2024-03-3038.P8306 【模板】字典树2024-03-3039.P1481 魔族密码2024-03-3040.P3128 [USACO15DEC] Max Flow P2024-04-0241.P5536 【XR-3】核心城市2024-04-0242.P5836 [USACO19DEC] Milk Visits S2024-04-0243.P3384 【模板】重链剖分/树链剖分2024-04-0344.P5960 【模板】差分约束2024-04-0445.P7771 【模板】欧拉路径2024-04-0546.六度分离2024-04-0847.整数区间2024-04-0848.F. Alex's whims2024-04-1249.J. 上学2024-05-1450.Game on Tree2024-05-1451.E. We Need More Bosses2024-05-1552.B. Omkar and Heavenly Tree2024-05-1653.B. Mahmoud and Ehab and the bipartiteness2024-05-1654.P1668 [USACO04DEC] Cleaning Shifts S2024-05-1755.P6154 游走2024-05-2056.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2457.P10298 [CCC 2024 S4] Painting Roads2024-05-2458.P9650 [SNCPC2019] Escape Plan2024-05-2759.P9327 [CCC 2023 S4] Minimum Cost Roads2024-05-2960.P9026 [CCC2021 S4] Daily Commute2024-05-2961.P8724 [蓝桥杯 2020 省 AB3] 限高杆2024-06-03

62.P4878 [USACO05DEC] Layout G2024-06-04

63.P5663 [CSP-J2019] 加工零件2024-06-0464.P2731 [USACO3.3] 骑马修栅栏 Riding the Fences2024-06-0865.I. Disks2024-06-1766.P1351 [NOIP2014 提高组] 联合权值2024-06-2667.B. Time Travel2024-06-2968.F. Minimum Maximum Distance2024-07-1469.A. Book2024-07-1970.P1407 [国家集训队] 稳定婚姻2024-07-2071.P1991 无线通讯网2024-07-2372.P4047 [JSOI2010] 部落划分2024-07-2373.P3275 [SCOI2011] 糖果2024-07-2374.P1989 无向图三元环计数2024-07-2775.P1967 [NOIP2013 提高组] 货车运输2024-07-3176.D. Vitaly and Cycle2024-08-0177.P10838 『FLA - I』庭中有奇树2024-08-0778.P9751 [CSP-J 2023] 旅游巴士2024-08-1179.D. Colored Portals2024-08-21

收起

原题链接

大概思路

我们已知一组不等式的解可以通过建边然后求最短路/最长路来得出
而这里要求 $D_n-D_1$ 的最大值，所以我们要求最短路。

补充

为什么要求最短路？
对于任何一组不等式，我们都可以写成 $a_i-b_i\le c_i$ 建边含义
假设 $D_n-D_1$ 有最大值，那么通过这组不等式的组合，一定能得出若干个形似 $D_n-D_1\le k_j$ 的式子，此时便可以得出 $max(D_n-D_1)$ 就等于 $min(k_j)$

细节

1. $D_i\ge D_{i-1}(i>1)\to D_{i-1}-D_i\le 0$
   2.要判断每条边（即每个约束）是否都满足，所以要建立超级源点，判断是否存在负环，不能以点1来判断是否存在负环，因为图不保证联通

code

#include<bits/stdc++.h>
using namespace std;
const int inf=2e9;

int dis[1010];

struct node
{
    int to,len;
};
vector<node> G[1010];
int n;
bool run(int start)
{
    int vis[1010]={0};
    bool in_q[1010]={0};
    fill(dis,dis+1004,inf);
    queue<int>  q;
    q.push(start);
    in_q[start]=1;
    dis[start]=0;

    while(q.size())
    {
        int now=q.front();
        q.pop();
        in_q[now]=0;
        vis[now]++;
        if(vis[now]>=n)
        {
            return 0;
        }
        for(auto next:G[now])
        {
            int to=next.to,len=next.len;
            if(dis[to]>dis[now]+len)
            {
                dis[to]=dis[now]+len;
                if(!in_q[to])
                {
                    q.push(to);
                }
            }
        }
    }
    return 1;
}
int main()
{
    int l,d;
    cin>>n>>l>>d;

    for(int i=1;i<=n;i++) G[0].push_back({i,0});
    for(int i=2;i<=n;i++) G[i].push_back({i-1,0});
    for(int i=1;i<=l;i++)
    {
        int a,b,c;
        cin>>a>>b>>c;
        G[a].push_back({b,c});
    }

    for(int i=1;i<=d;i++)
    {
        int a,b,c;
        cin>>a>>b>>c;
        G[b].push_back({a,-c});
    }



    if(!run(0)) puts("-1");
    else
    {
        run(1);
        if(dis[n]==inf) puts("-2");
        else cout<<dis[n]<<endl;
    }

    //for(int i=1;i<=n;i++) cout<<dis[i]<<" ";
    return 0;
}