POJ 3268 Silver Cow Party （最短路徑）

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

Http

POJ:https://vjudge.net/problem/POJ-3268

Source

最短路徑

題目大意

在一個有向圖中，求所有點都走到一個點再走回來的最短距離中的最大值

解決思路

我們知道單源最短路的求法，即從一個點走到其他點，那么我們只要把有向圖中的邊反過來求一遍就是從其他點走到一個點的最短距離
這里我們用spfa解決

代碼

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<vector> #include<queue> using namespace std;const int maxN=1001; const int inf=2147483647;class Edge { public:int v,w; };int n,m,X; vector<Edge> E1[maxN]; vector<Edge> E2[maxN]; queue<int> Q; bool inqueue[maxN]; int Dist1[maxN]; int Dist2[maxN];int main() {scanf("%d%d%d",&n,&m,&X);for (int i=1;i<=m;i++){int u,v,w;scanf("%d%d%D",&u,&v,&w);E1[u].push_back((Edge){v,w});//存正圖E2[v].push_back((Edge){u,w});//存反圖}memset(Dist1,127,sizeof(Dist1));//第一遍spfamemset(inqueue,0,sizeof(inqueue));Dist1[X]=0;inqueue[X]=1;while (!Q.empty())Q.pop();Q.push(X);do{int u=Q.front();Q.pop();inqueue[u]=0;for (int i=0;i<E1[u].size();i++){int v=E1[u][i].v;int w=E1[u][i].w;if (Dist1[v]>Dist1[u]+w){Dist1[v]=Dist1[u]+w;if (inqueue[v]==0){Q.push(v);inqueue[v]=1;}}}}while (!Q.empty());memset(Dist2,127,sizeof(Dist2));//第二遍spfamemset(inqueue,0,sizeof(inqueue));Dist2[X]=0;inqueue[X]=1;Q.push(X);do{int u=Q.front();Q.pop();inqueue[u]=0;for (int i=0;i<E2[u].size();i++){int v=E2[u][i].v;int w=E2[u][i].w;if (Dist2[v]>Dist2[u]+w){Dist2[v]=Dist2[u]+w;if (inqueue[v]==0){Q.push(v);inqueue[v]=1;}}}}while (!Q.empty());int Ans=0;for (int i=1;i<=n;i++)Ans=max(Ans,Dist1[i]+Dist2[i]);//統計最大值cout<<Ans<<endl;return 0; }

轉載于:https://www.cnblogs.com/SYCstudio/p/7225202.html

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