Silver Cow Party
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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.
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.
求两次最短路，第一次求x到其余各点的最短路，第二次求各点到x的最短路。前者易于解决，直接应用spfa或其他最短路算法即可，后者要先将邻接矩阵转置再执行最短路算法。
为什么进行矩阵转置？比如u（u ！= x）到x的最短路为<u,v1>,<v1,v2>,<v2,v3>,...,<vi, x>，这条路径在转置邻接矩阵后变成<x,vi>,...,<v3,v2>,<v2, v1>,<v1,u>.于是乎，在转置邻接矩阵后，执行最短路算法求出x到u的最短路<x,vi>,...,<v3,v2>,<v2, v1>,<v1,u>即可得到转置前u到x的最短路。
#include <iostream>
#include <deque>
#include <cstdio>
#include <cstring>
#include <algorithm> using namespace std; const int MAXV = ;
const int inf = 0x3f3f3f3f;
int t[MAXV][MAXV], d1[MAXV], d2[MAXV];
int que[MAXV<<];
bool in[MAXV];
int n, m, x; void spfa(int * d)
{
memset(in, false, sizeof(in));
memset(d + , inf, sizeof(int) * n);//memset(d, inf, sizeof(d)) if wrong
d[x] = ;
int tail = -;
que[++tail] = x;
in[x] = true;
while(tail != -){
int cur = que[tail];
tail--;
in[cur] = false;
for(int i = ; i <= n; i++){
if(d[cur] + t[cur][i] < d[i]){
d[i] = d[cur] + t[cur][i];
if(in[i] == false){
que[++tail] = i;
in[i] = true;
}
}
}
}
} void tran()
{
int i, j;
for(i = ; i <= n; i++){
for(j = ; j <= i; j++){
swap(t[i][j], t[j][i]);
}
}
} int main()
{
while(scanf("%d %d %d", &n, &m, &x) != EOF){
memset(t, inf, sizeof(t));
while(m--){
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
t[a][b] = c;
}
spfa(d1);
tran();
spfa(d2);
int ans = -;
for(int i = ; i <= n; i++){
if(d1[i] != inf && d2[i] != inf)
ans = max(ans, d1[i] + d2[i]);
}
printf("%d\n", ans);
}
return ;
}