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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P1491 集合位置

P1491 集合位置

次短路。

将最短路上的每一条边删除再做最短路得到的最小值即次短路。

查看代码
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#include <cstdio>
#include <cmath>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
typedef pair<double, int> PDI;
const double inf = 4e5 + 10;
const int N = 210, M = 8e4 + 10;
int n, m, pre[N];
PII p[N];
int idx = -1, hd[N], nxt[M], edg[M];
double d[N], wt[M];
double dijkstra(bool op)
{
for (int i = 1; i <= n; i++)
d[i] = inf;
d[1] = 0;
priority_queue<PDI, vector<PDI>, greater<PDI>> q;
q.push({0, 1});
while (!q.empty())
{
int t = q.top().second;
q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
if (d[t] + wt[i] < d[edg[i]])
{
d[edg[i]] = d[t] + wt[i];
if (op)
pre[edg[i]] = i;
q.push({d[edg[i]], edg[i]});
}
}
return d[n];
}
double dis(int x, int y)
{
int dx = abs(p[x].first - p[y].first), dy = abs(p[x].second - p[y].second);
return sqrt(dx * dx + dy * dy);
}
void add(int a, int b, double c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1; i <= n; i++)
scanf("%d%d", &p[i].first, &p[i].second);
for (int a, b; m; m--)
{
scanf("%d%d", &a, &b);
double c = dis(a, b);
add(a, b, c);
add(b, a, c);
}
dijkstra(true);
double res = inf;
for (int i = n; i != 1; i = edg[pre[i] ^ 1])
{
double tmp = wt[pre[i]];
wt[pre[i]] = inf;
res = min(res, dijkstra(false));
wt[pre[i]] = tmp;
}
if (res == inf)
puts("-1");
else
printf("%.2lf", res);
return 0;
}