题意:求起点1到终点n的次短路。
方法一:用两次dijkstra分别求出起点到所有点的最短距离d1[1...n]和所有点到终点的最短距离d2[1...n]。然后枚举每条边uv,如果d1[u] + uv + d2[v] 不等于最短路,且小于当前答案,更新即可。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
struct edg{
int v, nxt;
ll d;
}path[maxn * 2];
int pre[maxn], tot;
ll d1[maxn], d2[maxn];
void add(int u, int v, ll w) {
path[tot].v = v;
path[tot].d = w;
path[tot].nxt = pre[u];
pre[u] = tot++;
}
struct node{
int v;
ll dist;
bool operator < (const node &b) const {
return dist > b.dist;
}
};
int n, m;
bool vis[maxn];
void dijk(int s, ll d[]) {
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n; ++i) {
d[i] = 0x3f3f3f3f3f3f3f3f;//不要用 LONG_LONG_MAX !!!!
}
d[s] = 0;
priority_queue<node> que;
node p, q;
p.dist = 0;
p.v = s;
que.push(p);
while (!que.empty()) {
p = que.top();
que.pop();
if (vis[p.v]) {
continue;
}
vis[p.v] = true;
for (int i = pre[p.v]; ~i; i = path[i].nxt) {
if (!vis[path[i].v] && d[p.v] + path[i].d < d[path[i].v]) {
d[path[i].v] = d[p.v] + path[i].d;
q.v = path[i].v;
q.dist = d[path[i].v];
que.push(q);
}
}
}
}
int main(){
int t, u, v;
ll w;
scanf("%d", &t);
while (t--) {
tot = 0;
memset(pre, -1, sizeof(pre));
scanf("%d%d", &n, &m);
while (m--) {
scanf("%d%d%lld", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
dijk(1, d1);
dijk(n, d2);
ll mn = d1[n];
ll ans = LONG_LONG_MAX;
for (int i = 1; i <= n; ++i) {
for (int j = pre[i]; ~j; j = path[j].nxt) {
ll now = d1[i] + path[j].d + d2[path[j].v];
if (now > mn && now < ans) {
ans = now;
}
}
}
printf("%lld\n", ans);
}
return 0;
}
方法二:做一次dijkstra,dist数组改为两维,dist[i][0]记录到i的最短路,dist[i][1]记录到i的次短路。松弛的时候分别考虑最短和次短两种情况更新。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 100005;
struct edg{
int v, nxt;
ll d;
}path[maxn * 2];
int pre[maxn], tot;
ll dist[maxn][2];
void add(int u, int v, ll w) {
path[tot].v = v;
path[tot].d = w;
path[tot].nxt = pre[u];
pre[u] = tot++;
}
struct node{
int v, flag; //0表示最短路,1表示次短路
ll dist;
bool operator < (const node &b) const {
return dist > b.dist;
}
};
int n, m;
bool vis[maxn][2];
void dijk(int s) {
for (int i = 1; i <= n; ++i) {
vis[i][0] = vis[i][1] = false;
dist[i][0] = dist[i][1] = 0x3f3f3f3f3f3f3f3f;//不要用 LONG_LONG_MAX !!!!
}
dist[s][0] = 0;
priority_queue<node> que;
node p, q;
p.dist = 0;
p.v = s;
p.flag = 0;
que.push(p);
while (!que.empty()) {
p = que.top();
que.pop();
if (vis[p.v][p.flag]) {
continue;
}
vis[p.v][p.flag] = true;
for (int i = pre[p.v]; ~i; i = path[i].nxt) {
if (!vis[path[i].v][0] && p.dist + path[i].d < dist[path[i].v][0]) {
if (dist[path[i].v][0] != 0x3f3f3f3f3f3f3f3f) {
dist[path[i].v][1] = dist[path[i].v][0];
q.dist = dist[path[i].v][0];
q.flag = 1;
q.v = path[i].v;
que.push(q);
}
dist[path[i].v][0] = p.dist + path[i].d;
q.dist = dist[path[i].v][0];
q.flag = 0;
q.v = path[i].v;
que.push(q);
} else if (!vis[path[i].v][1] && p.dist + path[i].d < dist[path[i].v][1]) {
dist[path[i].v][1] = p.dist + path[i].d;
q.dist = dist[path[i].v][1];
q.flag = 1;
q.v = path[i].v;
que.push(q);
}
}
}
}
int main(){
int t, u, v;
ll w;
scanf("%d", &t);
while (t--) {
tot = 0;
memset(pre, -1, sizeof(pre));
scanf("%d%d", &n, &m);
while (m--) {
scanf("%d%d%lld", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
dijk(1);
printf("%lld\n", dist[n][1]);
}
return 0;
}