POJ 3268

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Silver Cow Party

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 23404 Accepted: 10678

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 T_i (1 ≤T_i ≤ 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: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

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

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

题意：

有向图，问所有点到某一点X来回的最短距离中最长的距离是多少。（有点拗口）

题目分析：

一开始想的是先从终点X跑一遍Dijkstra或者SPFA，Floyd指定超时，数量级达到了10^3。然后再从各点跑一遍Dijkstra或者SPFA，两次加和求最大值即可，但是超时了。后来灵光一现，想到可以倒着跑一遍，这样只需跑两遍Dijkstra或者SPFA即可，

但在存距离的时候要存在两个数组中了。

AC代码：

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <climits>
using namespace std;
int N, M, X;
const int MAXN=1e3+10;
const int maxx=INT_MAX;
int data1[MAXN][MAXN], data2[MAXN][MAXN], Distance[MAXN], z[MAXN];
bool pin[MAXN], collect1[MAXN][MAXN], collect2[MAXN][MAXN];
void start()
{
for(int i=1; i<=N; i++)
{
for(int j=1; j<=N; j++)
{
if(i==j)
{
data1[i][j]=0;
data2[i][j]=0;
}
else
{
data1[i][j]=maxx;
data2[i][j]=maxx;
}
}
}
memset(collect1,false,sizeof(collect1));
memset(collect2,false,sizeof(collect2));
memset(pin,false,sizeof(pin));
memset(z,0,sizeof(z));
}
bool cmp(int a, int b)
{
return a>b;
}
int zhao()
{
int a=-1, b=maxx;
for(int i=1; i<=N; i++)
{
if(!pin[i]&&Distance[i]<b)
{
a=i;
b=Distance[i];
}
}
return a;
}
void Dijkstra(int x, bool collect[][MAXN], int data[][MAXN])
{
Distance[x]=0;
pin[x]=true;
while(1)
{
int m=zhao();
if(m==-1)
break;
pin[m]=true;
for(int i=1; i<=N; i++)
{
if(collect[m][i]&&!pin[i])
Distance[i]=min(Distance[i],Distance[m]+data[m][i]);
}
}
}
void SPFA(int x, bool collect[][MAXN], int data[][MAXN])
{
queue<int>Q;
Distance[x]=0;
Q.push(x);
pin[x]=true;
while(!Q.empty())
{
int m=Q.front();
Q.pop();
pin[m]=false;
for(int i=1; i<=N; i++)
{
if(collect[m][i]&&Distance[i]>Distance[m]+data[m][i])
{
Distance[i]=Distance[m]+data[m][i];
if(!pin[i])
{
Q.push(i);
pin[i]=true;
}
}
}
}
}
int main()
{
while(cin>>N>>M>>X)
{
start();
int a, b, c;
while(M--)
{
cin>>a>>b>>c;
if(data1[a][b]>c)
{
collect1[a][b]=true;///分别存在两个不同数组中
collect2[b][a]=true;
data1[a][b]=c;
data2[b][a]=c;
}
}
for(int i=1; i<=N; i++)
Distance[i]=data1[X][i];
Dijkstra(X,collect1,data1);///Dijkstra算法的Distance数组需要提前赋值，否则在找Distance最小时出错，而SPFA算法则不用，直接赋值为maxx即可

for(int i=1; i<=N; i++)
z[i]+=Distance[i];
for(int j=1; j<=N; j++)
Distance[j]=data2[X][j];
memset(pin,false,sizeof(pin));
Dijkstra(X,collect2,data2);
for(int i=1; i<=N; i++)
z[i]+=Distance[i];
sort(z+1,z+N+1,cmp);
cout<<z[1]<<endl;
}
return 0;
}

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