次最短路的思想:
起點s, 終點t,中間點u,次最短路可以通過到u的最短路+u到t的次短路,或者t到u 的最短路+u到t的最短路。
例題:Roadblocks
代碼:
#include<iostream>
#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
const int maxn = 200100;
const int INF = 1e9;
int h[maxn], to[maxn], w[maxn], nex[maxn], tot = 0;
void add(int u, int v, int vv){
to[tot] = v;
w[tot] = vv;
nex[tot] = h[u];
h[u] = tot++;
}
int dis[5010], dis1[5010];
int n, m;
typedef pair<int, int> PP;
int dijstra(){
for(int i = 1; i <= n; i++){
dis[i] = INF;
dis1[i] = INF;
}
dis[1] = 0;
priority_queue<PP, vector<PP>, greater<PP> >q;
q.push({0, 1});
while(!q.empty()){
int u = q.top().second;
int d = q.top().first;
q.pop();
if(dis1[u] < d)continue;
for(int i = h[u]; i != -1; i = nex[i]){
int val = w[i], v = to[i];
int d2 = d + val;
if(d2 < dis[v]){
dis[v] = d2;
q.push({d2, v});
}
if(d2 > dis[v] && d2 < dis1[v]){
dis1[v] = d2;
q.push({d2, v});
}
}
}
return dis1[n];
}
int main(){
scanf("%d %d", &n, &m);
memset(h, -1, sizeof h);
for(int i = 1; i <= m; i++){
int u, v, w;
scanf("%d %d %d", &u, &v, &w);
add(u, v, w);
add(v, u, w);
}
printf("%d\n", dijstra());
return 0;
}