luogu P2865 [USACO06NOV]路障Roadblocks

analysis

第2短路,新模型

核心思想是在最短路更新的时候同时带上次短路更新

这是一定可行的!

讨论如下:

$$\begin{gathered} 对于边u->v \\ 设di{s}_{(x,0)}为起点x的最短路长度,设di{s}_{(x,1)}为起点x的次短路长度 \end{gathered}$$

$$\begin{gathered} 先考虑di{s}_{(u,0)} \\ 1.如果di{s}_{(u,0)}+edge[i].w<di{s}_{(v,0)},说明v的最短路被更新 \\ 那么v的次短路一起更新 \\ 2.如果di{s}_{(u,0)}+edge[i].w<di{s}_{(v,1)}且di{s}_{(u,0)}+edge[i].w>di{s}_{(v,0)} \\ 说明只能够更新到v的次短路,更新即可 \\ 3.di{s}_{(u,0)}+edge[i].w>di{s}_{(v,1)}时,不做考虑 \end{gathered}$$

$$\begin{gathered} 再考虑di{s}_{(u,1)} \\ 1.如果di{s}_{(u,1)}+edge[i].w<di{s}_{(v,0)} \\ 说明v的最短路被u的次短路更新 \\ 而di{s}_{(u,1)}>di{s}_{(u,0)} \\ 那么u的最短路更新一定最优 \\ 言下之意就是已经被更新过了,这种情况不会出现 \\ 2.如果di{s}_{(u,1)}+edge[i].w<di{s}_{(v,1)}且di{s}_{(u,1)}+edge[i].w>di{s}_{(v,0)} \\ 说明只能够更新到v的次短路,更新即可 \\ 3.di{s}_{(u,1)}+edge[i].w>di{s}_{(v,1)}时,不做考虑 \end{gathered}$$

$综上,有3种情况需要处理(详见代码里的3个if)$

code

#include
using namespace std;
#define loop(i,start,end) for(register int i=start;i<=end;++i)
#define clean(arry,num) memset(arry,num,sizeof(arry))
#define anti_loop(i,start,end) for(register int i=start;i>=end;--i)
#define ll long long
template<typename T>void read(T &x){
	x=0;char r=getchar();T neg=1;
	while(r>'9'||r<'0'){if(r=='-')neg=-1;r=getchar();}
	while(r>='0'&&r<='9'){x=(x<<1)+(x<<3)+r-'0';r=getchar();}
	x*=neg;
}
int n,m;
const int maxn=5000+10;
const int maxm=100000+10;
struct node{
	int e;
	ll w;
	int nxt;
}edge[maxm<<1];
int head[maxn];
int cnt=0;
inline void addl(int u,int v,ll w){
	edge[cnt].e=v;
	edge[cnt].w=w;
	edge[cnt].nxt=head[u];
	head[u]=cnt++;
}
ll dis[maxn][2];
struct point{
	int pos;
	ll dis;
	point():pos(0),dis(0){}
	point(int pos,ll dis):pos(pos),dis(dis){}
	friend bool operator<(point a,point b){
		return a.dis>b.dis;
	}
};
priority_queue<point>q;
void dijkstra(){
	clean(dis,0x3f);
	q.push(point(1,dis[1][0]));
	dis[1][0]=0;
	//dis[1]
	while(q.empty()==false){
		int f=q.top().pos;
		q.pop();
		for(int i=head[f];i!=-1;i=edge[i].nxt){
			int v=edge[i].e;
			if(dis[v][0]>dis[f][0]+edge[i].w){
				dis[v][1]=dis[v][0];
				dis[v][0]=dis[f][0]+edge[i].w;
				q.push(point(v,dis[v][0]));
			}
			if(dis[v][1]>dis[f][0]+edge[i].w&&dis[f][0]+edge[i].w>dis[v][0]){
				dis[v][1]=dis[f][0]+edge[i].w;
				q.push(point(v,dis[v][1]));
			}
			if(dis[v][1]>dis[f][1]+edge[i].w&&dis[f][1]+edge[i].w>dis[v][0]){
				dis[v][1]=dis[f][1]+edge[i].w;
				q.push(point(v,dis[v][1]));
			}
		}
	}
	printf("%lld\n",dis[n][1]);
}
int main(){
	#ifndef ONLINE_JUDGE
	freopen("datain.txt","r",stdin);
	#endif
	clean(head,-1);
	read(n);
	read(m);
	loop(i,1,m){
		int ai,bi,di;
		read(ai);
		read(bi);
		read(di);
		addl(ai,bi,(ll)di);
		addl(bi,ai,(ll)di);
	}
	dijkstra();
	return 0;
}