次短路

模型建立

一般题目中会明确告诉你要求第二短路或者次短路

思路

主要思路一共有两种，一种为记录路径然后删边，另一种是开一个dis2记录次短路.

暴力删边

好像没什么可讲的
以洛谷P1491 集合位置为例

直接上代码吧..

#include <bits/stdc++.h>
#define ll long long
#define N 100010
#define M 210

using namespace std;
int n, m, add_edge;
bool bian[M][M];
double px[M], py[M], dis[M];
int head[M * M * 2], from[M], vis[M];
struct CCC {
  int next, to;
  double dis;
} edge[M * M * 2];
struct node {
  int x;
  double dis;
  bool operator < (const node &b) const {
    return dis > b.dis;
  }
};

int read() {
  int s = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
  return f ? -s : s;
}

double distant(double x, double y, double a, double b) {
  return sqrt((x - a) * (x - a) + (y - b) * (y - b));
}

void add(int from, int to, double dis) {
  edge[++add_edge].next = head[from];
  edge[add_edge].to = to;
  edge[add_edge].dis = dis;
  head[from] = add_edge;
}

void dijkstra(int sy) {
  for (int i = 1; i <= n; i++) dis[i] = 444444444444.00;
  memset(vis, 0, sizeof vis);
  priority_queue<node> q;
  q.push((node) {
    1, 0
  }), dis[1] = 0;
  while (!q.empty()) {
    node fr = q.top();
    q.pop();
    int x = fr.x;
    if (vis[x]) continue;
    vis[x] = 1;
    for (int i = head[x]; i; i = edge[i].next) {
      int to = edge[i].to;
      if (bian[x][to] || bian[to][x]) continue;
      if (!vis[to] && dis[to] > dis[x] + edge[i].dis) {
        dis[to] = dis[x] + edge[i].dis;
        if (sy) from[to] = x;
        q.push((node) {
          to, dis[to]
        });
      }
    }
  }
}

int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++)
    px[i] = read(), py[i] = read();
  for (int i = 1, x, y; i <= m; i++) {
    x = read(), y = read();
    double d = distant(px[x], py[x], px[y], py[y]);
    add(x, y, d), add(y, x, d);
  }
  dijkstra(1);
  double an = dis[n];
  int sy = n;
  double ans = 444444444444.00;
  while (sy != 1) {
    bian[from[sy]][sy] = 1, bian[sy][from[sy]] = 1;
    dijkstra(0);
    if (dis[n] > an)
      ans = min(ans, dis[n]);
    bian[from[sy]][sy] = 0, bian[sy][from[sy]] = 0;
    sy = from[sy];
  }
  if (ans == 444444444444.00) puts("-1");
  else printf("%.2lf\n", ans);
}

the other

以洛谷P2865 [USACO06NOV]Roadblocks G为例
这个题以为有重边，然后暴力删边操作之后只有90分就很难受.

在用\(dijkstra\)时候，就直接更新次短路，用\(dis[]\)代表最短路，用\(dis2[]\)代表次短路，
然后在松弛的时候同时更新两个数组，要判断三个条件
（\(u\)是当前考虑的点，\(v\)是与\(u\)有边相连的点，\(d(u,v)\)表示从\(u\)到\(v\)的边长）

\(1.\)如果\(dis[v]>dis[u]+d(u,v)\)，则更新\(dis[v]\)
\(2.\)如果\(dis[v]<dis[u]+d(u,v)\)（不能取等，否则\(dis2[v]\)和\(dis[v]\)可能相等）
且\(dis2[v]>dis[u]+d(u,v)\)，则更新\(dis2[v]\)
\(3.\)如果\(dis2[v]>dis2[u]+d(u,v)\)，则更新\(dis2[v]\)（显然，如果2成立，
更新后\(dis2[v]=dis[u]+d(u,v)<dis2[u]+d(u,v)\)，即\(3\)一定不成立）

如果上述三个条件中有任意一个成立，则将v入队。

code

#include <bits/stdc++.h>
#define ll long long
#define N 100010
#define M 5010

using namespace std;
int n, m;
int head[N << 1], add_edge, nx, ny, ndis;
int dis[M], dis2[M], vis[M], from[M], bian[M][M];
struct CCC {
  int next, to, dis;
} edge[N << 1];
struct node {
  int x, dis;
  bool operator < (const node &b) const {
    return dis > b.dis;
  }
};

int read() {
  int s = 0, f = 0; char ch = getchar();
  while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
  return f ? -s : s;
}

void add(int from, int to, int dis) {
  edge[++add_edge].next = head[from];
  edge[add_edge].to = to;
  edge[add_edge].dis = dis;
  head[from] = add_edge;
}

void dijkstra(int sy) {
  memset(dis, 0x3f, sizeof dis);
  memset(dis2, 0x3f, sizeof dis2);
  priority_queue<node> q;
  q.push((node) {1, 0}), dis[1] = 0;
  while (!q.empty()) {
    node fr = q.top();
    q.pop();
    int x = fr.x;
    for (int i = head[x]; i; i = edge[i].next) {
      int to = edge[i].to;
      if (dis[to] >= dis[x] + edge[i].dis) {
        dis[to] = dis[x] + edge[i].dis;
        q.push((node) {to, dis[to]});
      } else if (dis2[to] > dis[x] + edge[i].dis) {
        dis2[to] = dis[x] + edge[i].dis;
        q.push((node) {to, dis[to]});
      }
      if (dis2[to] > dis2[x] + edge[i].dis)
        dis2[to] = dis2[x] + edge[i].dis;
    }
  }
}

int main() {
  n = read(), m = read();
  for (int i = 1, x, y, d; i <= m; i++) {
    x = read(), y = read(), d = read();
    add(x, y, d), add(y, x, d);
    if (!bian[x][y]) bian[x][y] = d, bian[y][x] = d;
    else bian[x][y] = min(bian[x][y], d), bian[y][x] = bian[x][y];
  }
  dijkstra(1);
  cout << dis2[n] << "\n";
}