Silver Cow Party ![分享至QQ空间]()
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Total Submit: 58 Accepted: 28
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 farmBi,
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
Hint
Source
思路：对原图求一次最短路，再对反图进行一次最短路，然后找最求解即可.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
using namespace std;
const int maxn = 1100;
const int maxm = 101000;
const int INF = 0x7fffffff;
vector<pair<int, int> > g[maxn];
vector<pair<int, int> > rg[maxn];//reverse;
int gn, gm, x;
bool inque[maxn];
queue<int> Q;
int d[maxn];//保存原图最短径.
int rd[maxn];//保存反图的最短路径值.
void spfa2(int s) {
int i;
memset(inque, false, sizeof(inque));
for(i = 1; i < maxn; i++) rd[i] = INF;
rd[s] = 0;
while(!Q.empty()) Q.pop();
Q.push(s);
inque[s] = true;
while(!Q.empty()) {
int u = Q.front();
Q.pop();
for(i = 0; i < (int)rg[u].size(); i++) {
int t = rg[u][i].first;
if( rd[u] != INF && rd[u] + rg[u][i].second < rd[t]) {
rd[t] = rd[u] + rg[u][i].second;
if(!inque[t]) {
inque[t] = true;
Q.push(t);
}
}
}
inque[u] = false;
}
}
void work(const int s) {
int i;
int maxv = -1;
for(i = 1; i <= gn; i++) {
if(i != s) {
if(d[i] + rd[i] > maxv) {
maxv = d[i] + rd[i];
}
}
}
printf("%d\n", maxv);
}
void spfa1(int s) {
int i;
memset(inque, false, sizeof(inque));
for(i = 1; i < maxn; i++) d[i] = INF;
d[s] = 0;
while(!Q.empty()) Q.pop();
Q.push(s);
inque[s] = true;
while(!Q.empty()) {
int u = Q.front();
Q.pop();
for(i = 0; i < (int)g[u].size(); i++) {
int t = g[u][i].first;
if(d[u] != INF && d[u] + g[u][i].second < d[t]) {
d[t] = d[u] + g[u][i].second;
if(!inque[t]) {
inque[t] = true;
Q.push(t);
}
}
}
inque[u] = false;
}
}
int main()
{
int i;
int x, y, w;
int start;
pair<int, int> t;
while(scanf("%d%d%d", &gn, &gm, &start) != EOF) {
for(i = 1; i <= gn; i++) {
g[i].clear();
rg[i].clear();
}
for(i = 0; i < gm; i++) {
scanf("%d%d%d", &x, &y, &w);
t.first = y;
t.second = w;
g[x].push_back(t);
t.first = x;
t.second = w;
rg[y].push_back(t);
}
spfa1(start);
spfa2(start);
work(start);
}
return 0;
}