【USACO3.2.6】香甜的黄油 最短路径

按照题目意思，就是求： 从K点开始，到所有点的最短路径之和。

穷举K即可。  最短路算法，我用的是SPFA。

SPFA+SLF优化：

Compiling...
Compile: OK

Executing...
Test 1: TEST OK [0.005 secs, 3508 KB]
Test 2: TEST OK [0.003 secs, 3508 KB]
Test 3: TEST OK [0.005 secs, 3508 KB]
Test 4: TEST OK [0.005 secs, 3508 KB]
Test 5: TEST OK [0.011 secs, 3508 KB]
Test 6: TEST OK [0.022 secs, 3508 KB]
Test 7: TEST OK [0.049 secs, 3508 KB]
Test 8: TEST OK [0.051 secs, 3508 KB]
Test 9: TEST OK [0.078 secs, 3508 KB]
Test 10: TEST OK [0.084 secs, 3508 KB]

All tests OK.

/*
TASK:butter
LANG:C++
*/
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

int cows, pastures, paths;
const int savebuff = 50;
int a[500 + savebuff], f[1000];
bool vis[1000];
int ans = 0x7fffffff;
struct edge
{
int dot, dis;
edge *next;
edge()
{
dot = dis = -1;
next = NULL;
}
edge(int DOT_, int DIS_, edge *NEXT_)
{
dot = DOT_;
dis = DIS_;
next = NEXT_;
}
}*x[800 + savebuff];

inline void insert(int A, int B, int dis)
{
x[A] = new edge(B, dis, x[A]);
}
deque<int>q;
void SPFA(int s)
{
q.push_back(s);
memset(f, 60, sizeof(f));
memset(vis, false, sizeof(vis));
f[s] = 0;
vis[s] = true;
while (!q.empty())
{
int now = q.front();
vis[now] = false;
q.pop_front();
for (edge *i = x[now]; i; i = i -> next)
{
int will = i -> dot;
int cost = i -> dis;
if (f[now] + cost < f[will])
{
f[will] = f[now] + cost;
if (!vis[will])
{
if (!q.empty() && f[will] < f[q.front()]) q.push_front(will);
else q.push_back(will);
}
}
}
}
int tmpans = 0;
for (int i = 0; i != cows; ++ i) tmpans += f[a[i]];
if (tmpans < ans) ans = tmpans;
}

int main()
{
freopen("butter.in","r",stdin);
freopen("butter.out","w",stdout);
memset(x, NULL, sizeof(x));
scanf("%d%d%d", &cows, &pastures, &paths);
for (int i = 0; i != cows; ++ i) scanf("%d", a+i);
while (paths--)
{
int ta, tb, tc;
scanf("%d%d%d", &ta, &tb, &tc);
insert(ta, tb, tc);
insert(tb, ta, tc);
}
for (int i = 1; i <= pastures; ++ i) SPFA(i);
printf("%d\n", ans);
return 0;
}

==============

SPFA的LLL+SLF优化： 速度更慢，就不贴代码了。