P9650 [SNCPC2019] Escape Plan

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 Roads2024-05-24

58.P9650 [SNCPC2019] Escape Plan2024-05-27

59.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2024-06-0362.P4878 [USACO05DEC] Layout G2024-06-0463.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

开始

敲代码前问问自己：如何证明你的算法的正确性？

启发

我们一开始在节点1，假如有abcd四条边，且我们已知 从每条边出发到达终点的最短距离 $a<b<c<d$ 且 $d[1]=3$
请问我们要封 节点1 的哪条边？
答案显然是abc

那么好接下来我们走边d，假设到达了节点3，我们把节点3当成节点1，又回到了上述情景

一句话总结

我们要率先封住那些 使得当前节点到终点距离最小的边

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll maxs=2e18;
int e[100005];
int d[100005];
ll dis[100005];
struct node
{
    int to,len;
};
vector<node> G[100005];

struct fresh
{
    int pos,val;
    bool operator<(const fresh &b) const
    {
        return b.val<val;
    }
};

void solve()
{
    int n,m,k;
    cin>>n>>m>>k;

    for(int i=1;i<=k;i++) cin>>e[i];

    for(int i=1;i<=n;i++)
    {
        cin>>d[i];
        dis[i]=maxs;
    }

    for(int i=1;i<=m;i++)
    {
        int x,y,w;
        cin>>x>>y>>w;
        G[x].push_back({y,w});
        G[y].push_back({x,w});
    }

    priority_queue<fresh> q;
    for(int i=1;i<=k;i++)
    {
        q.push({e[i],0});
        d[e[i]]=0;
    }

    while(q.size())
    {
        int now=q.top().pos,val=q.top().val;
        q.pop();

        if(dis[now]<=val) continue;
        if(d[now])
        {
            d[now]--;
            continue;
        }
        dis[now]=val;
        for(auto next:G[now])
        {
            int to=next.to,len=next.len;
            if(dis[now]+len<dis[to])
            {
                q.push({to,len+dis[now]});
            }
        }
    }

    if(dis[1]!=maxs) cout<<dis[1]<<"\n";
    else cout<<"-1\n";

    for(int i=1;i<=n;i++)
    {
        G[i].clear();
    }
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}