LOJ1379限制性最路算法应用

来源：互联网发布：linux 多核cpu使用率编辑：程序博客网时间：2024/06/15 20:34

思路：题目求的是从S->T的路径中权值小于给定P值的最大值；这样的话我们就只能枚举假定必经过的边（枚举）<a,b>,

此时就是ans = dis<S->a> + w<a,b> + dis<b->T> 设纪录最终结果的变量是rec，如果rec < ans && ans <= P,那么rec = ans;

这里dis<S->a>显然由正向的单源最短路,那么dis<b->T>就是反向的单源最短路；

题目链接

/*限制性的最短路变种，求最短路中的最长边*//*两次spfa + dp*/#include <iostream>#include <algorithm>#include <string>#include <stack>#include <queue>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>using namespace std;const int maxn = 22222;const int maxe = 222222;const int inf = 1 << 30;#define FILL(a,b) memset(a,b,sizeof a)#define CLR(a) memset(a,0,sizeof a)template<class T> inline T Get_Max(const T&a,const T&b) {return a < b?b:a;}template<class T> inline T Get_Min(const T&a,const T&b) {return a < b?a:b;}struct Edge{int v,w,next;};Edge E1[maxe],E2[maxe];int head1[maxn],head2[maxn];int cnt1,cnt2,n,m,st,ed,p;inline void Addedge1(int u,int v,int w){E1[cnt1].v = v;E1[cnt1].w = w;E1[cnt1].next = head1[u];head1[u] = cnt1++;}inline void Addedge2(int u,int v,int w){E2[cnt2].v = v;E2[cnt2].w = w;E2[cnt2].next = head2[u];head2[u] = cnt2++;}int dis[2][maxn];bool vis[maxn];inline void spfa(int st,int ed,Edge *edge,int* dis,int *head){FILL(vis,false);fill(dis,dis + 2 + n,inf);// memset(dis,inf,sizeof dis);queue<int> que;que.push(st);dis[st] = 0;vis[st] = true;while(!que.empty()){int u = que.front();que.pop();vis[u] = false;for (int i = head[u];i != -1;i = edge[i].next){int v = edge[i].v;int w = edge[i].w;if (dis[u] + w < dis[v]){dis[v] = dis[u] + w;if (!vis[v]){vis[v] = true;que.push(v);}}}}// cout << "dis = " << dis[n] << endl;// printf("n->1 %d\n",dis[1]);// printf("1->n %d\n",dis[n]);}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);int kase;int iCase = 0;int u,v,w;scanf("%d",&kase);while(kase--){scanf("%d%d%d%d%d",&n,&m,&st,&ed,&p);cnt1 = cnt2 = 0;FILL(head1,-1);FILL(head2,-1);while(m--){scanf("%d%d%d",&u,&v,&w);Addedge1(u,v,w);Addedge2(v,u,w);}spfa(st,ed,E1,dis[0],head1);spfa(ed,st,E2,dis[1],head2);int ans = -1;for (int u = 1;u <= n;u++){for (int i = head1[u];i != -1;i = E1[i].next){int v = E1[i].v;int w = E1[i].w;if (dis[0][u] < inf && dis[1][v] < inf && dis[0][u] + dis[1][v] + w <= p){ans = Get_Max(ans,w);}}}printf("Case %d: %d\n",++iCase,ans);}return 0;}

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