Codeforces 96D Volleyball spfa

来源：互联网发布：网络层传输的数据单位编辑：程序博客网时间：2024/05/24 01:54

题目链接：点击打开链接

题意：

给定n个点m条边的无向图

起点、终点

下面m行表示边和边权

再下面n行表示每个点有一辆出租车，这辆出租车能开的最远距离和搭乘这辆车的费用

问到终点的最小费用

开始感觉复杂度太大不好下手，暴力出奇迹。。

Y一下即可得到 spfa套spfa

注意inf要足够大，__int64

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<math.h>#include<set>#include<queue>#include<map>#include<vector>using namespace std;#define N 1005#define inf 10000000000000#define ll __int64struct Edge{ll from, to, dis, nex;}edge[N<<1];ll head[N],edgenum;void add(ll u,ll v,ll d){Edge E = {u,v,d,head[u]};edge[edgenum] = E;head[u] = edgenum++;}ll cost[N], far[N];vector<ll>G[N];ll dis[N];bool vis[N], inq[N];ll n, m, st, en;void dou(ll x){if(vis[x])return;vis[x] = 1;for(ll i = 1; i <= n; i++)dis[i] = inf, inq[i] = 0;dis[x] = 0;queue<ll>q; q.push(x);while(!q.empty()){ll u = q.front(); q.pop(); inq[u] = 0;for(ll i = head[u]; ~i; i = edge[i].nex){ll v = edge[i].to;if(dis[u]+edge[i].dis<=far[x] && dis[v]>dis[u]+edge[i].dis){dis[v] = edge[i].dis+dis[u];if(!inq[v])inq[v] = 1, q.push(v);}}}for(ll i = 1; i <= n; i++)if(!vis[i] && dis[i]<=far[x])G[x].push_back(i);}ll ned[N];bool hehe[N];ll bfs(){for(ll i = 1; i <= n; i++)ned[i] = inf, hehe[i] = 0;ned[st] = 0;queue<ll>q;q.push(st);hehe[en] = 1;while(!q.empty()){ll u = q.front(); q.pop(); hehe[u] = 0;dou(u);for(ll i = 0; i < G[u].size(); i++){ll v = G[u][i];if(ned[v]>ned[u]+cost[u]){ned[v] = ned[u]+cost[u];if(!hehe[v])hehe[v]=1,q.push(v);}}}if(ned[en]==inf)return -1;return ned[en];}void init(){memset(vis, 0, sizeof vis);for(ll i = 1; i <= n; i++)G[i].clear();memset(head,-1,sizeof head); edgenum = 0;}int main(){ll i, j, u, v, d;while(cin>>n>>m){init();cin>>st>>en;while(m--){cin>>u>>v>>d;add(u,v,d);add(v,u,d);}for(i=1;i<=n;i++)cin>>far[i]>>cost[i];cout<<bfs()<<endl;}return 0;}

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