POJ 3268

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define N 1200
#define inf 0x3f3f3f3f
struct eg
{
int pos, val;
eg(int pp, int vv) :pos(pp), val(vv){}
eg(){}
};
vector<eg>g1
;
vector<eg>g2
;
int vis
, d1
, d2
;
void spfa1(int dest)
{
memset(d1, inf, sizeof(d1));
memset(vis, 0, sizeof(vis));
vis[dest] = 1;
d1[dest] = 0;
queue<int>q;
q.push(dest);
while (!q.empty())
{
int x = q.front(); q.pop();
vis[x] = 0;
for (int i = 0; i < g1[x].size(); i++)
{
int v = g1[x][i].pos;
if (v == x)continue;
int cost = g1[x][i].val;
if (d1[v]>d1[x] + cost)
{
d1[v] = d1[x] + cost;
if (!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
}
void spfa2(int dest)
{
memset(d2, inf, sizeof(d2));
memset(vis, 0, sizeof(vis));
vis[dest] = 1;
d2[dest] = 0;
queue<int>q;
q.push(dest);
while (!q.empty())
{
int x = q.front(); q.pop();
vis[x] = 0;
for (int i = 0; i < g2[x].size(); i++)
{
int v = g2[x][i].pos;
if (v == x)continue;
int cost = g2[x][i].val;
if (d2[v]>d2[x] + cost)
{
d2[v] = d2[x] + cost;
if (!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
}
int main()
{
#ifdef CDZSC
freopen("in.txt", "r", stdin);
#endif
int n, m, des, x, y, v;
while (~scanf("%d%d%d", &n, &m, &des))
{
while (m--)
{
scanf("%d%d%d", &x, &y, &v);
g1[x].push_back(eg(y, v));
g2[y].push_back(eg(x, v));
}
spfa1(des); spfa2(des);
int ans = -inf;
for (int i = 1; i <= n; i++)
ans=max(ans,d1[i]+d2[i]);
printf("%d\n", ans);
}
return 0;
}

题意是给你一个点des作为终点，你需要求出所有的点到des的最短路，再求出des到所有点的最短路，求在这来回的路径中最长的那条路。开始大家都会很容易想到用Floyd去求任意点的最短路，但是O(n^3）明显1s不能够承受这么大的复杂度。换个思路，将矩阵转秩。只需要求2次最短路就可以了。以上为代码。以下是SPFA模板.

vector<eg>g
;
int vis
,d
;
void spfa(int dest)
{
memset(d, inf, sizeof(d));
memset(vis, 0, sizeof(vis));
vis[dest] = 1;
d[dest] = 0;
queue<int>q;
q.push(dest);
while (!q.empty())
{
int x = q.front(); q.pop();
vis[x] = 0;
for (int i = 0; i < g[x].size(); i++)
{
int v = g[x][i].pos;
if (v == x)continue;
int cost = g[x][i].val;
if (d[v]>d[x] + cost)
{
d[v] = d[x] + cost;
if (!vis[v])
{
q.push(v);
vis[v] = 1;
}
}
}
}
}

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15910		Accepted: 7248

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

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.
Source

USACO 2007 February Silver