【题解】LuoGu2865：[USACO06NOV]Roadblocks G

原题传送门
求严格次小最短路
直接用dijkstra求
令 $dis_{u,0}$为最短路， $dis_{u,1}$为次短路
初始化只有 $dis_{1,0}$是0，其他都是inf
对于一条边 $(u,v,l)路径长度为l$

• $若dis_{v,0}>dis_{u,0}+l，则dis_{v,0}=dis_{u,0}+l$
• $若dis_{v,0}<dis_{u,0}+l且dis_{v,1}>dis_{u,0}+l，则dis_{v,1}=dis_{u,0}+l$
• $若dis_{v,1}>dis_{u,1}+l，则dis_{v,1}=dis_{u,1}+l$

上述条件满足其一，则入优先队列

Code：

#include <bits/stdc++.h>
#define maxn 100010
using namespace std;
struct heap{
	int val, len;
	bool operator < (const heap &x) const{ return x.len < len; }
};
priority_queue <heap> q;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
int num, head[maxn], dis[maxn][2], vis[maxn], n, m;

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y, int z){ edge[++num] = (Edge){y, head[x], z}, head[x] = num; }

int main(){
	n = read(), m = read();
	for (int i = 1; i <= m; ++i){
		int x = read(), y = read(), z = read();
		addedge(x, y, z), addedge(y, x, z);
	}
	for (int i = 2; i <= n; ++i) dis[i][0] = dis[i][1] = 1e9;
	dis[1][1] = 1e9;
	q.push((heap){1, 0});
	while (!q.empty()){
		heap tmp = q.top(); q.pop();
		int u = tmp.val;
		for (int i = head[u]; i; i = edge[i].next){
			int v = edge[i].to, flag = 0;
			if (dis[v][0] > dis[u][0] + edge[i].len) dis[v][0] = dis[u][0] + edge[i].len, flag = 1;
			if (dis[v][1] > dis[u][0] + edge[i].len && dis[v][0] < dis[u][0] + edge[i].len) dis[v][1] = dis[u][0] + edge[i].len, flag = 1;
			if (dis[v][1] > dis[u][1] + edge[i].len) dis[v][1] = dis[u][1] + edge[i].len, flag = 1;
			if (flag) q.push((heap){v, dis[v][0]});
		}
	}
	printf("%d\n", dis[n][1]);
	return 0;
}