【题解】
裸的最短路。直接跑dijkstra即可。
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #define LL long long 5 #define rg register 6 #define N 200010 7 using namespace std; 8 int n,tot,last[N],dis[N],pos[N]; 9 struct edge{ 10 int to,pre,dis; 11 }e[N]; 12 struct heap{ 13 int poi,dis; 14 }h[N]; 15 inline int read(){ 16 int k=0,f=1; char c=getchar(); 17 while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar(); 18 while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar(); 19 return k*f; 20 } 21 inline void up(int x){ 22 int fa; 23 while((fa=(x>>1))&&h[fa].dis>h[x].dis) 24 swap(h[fa],h[x]),swap(pos[h[fa].poi],pos[h[x].poi]),x=fa; 25 } 26 inline void down(int x){ 27 int son; 28 while((son=(x<<1))<=tot){ 29 if(h[son].dis>h[son+1].dis&&son<tot) son++; 30 if(h[son].dis<h[x].dis) 31 swap(h[son],h[x]),swap(pos[h[son].poi],pos[h[x].poi]),x=son; 32 else return; 33 } 34 } 35 inline void dijkstra(int x){ 36 for(rg int i=1;i<=n;i++) dis[i]=1e9; 37 h[tot=pos[x]=1]=(heap){x,dis[x]=0}; 38 while(tot){ 39 int now=h[1].poi; h[1]=h[tot--]; if(tot) down(1); 40 for(rg int i=last[now],to;i;i=e[i].pre) 41 if(dis[to=e[i].to]>dis[now]+e[i].dis){ 42 dis[to]=dis[now]+e[i].dis; 43 if(!pos[to]) h[pos[to]=++tot]=(heap){to,dis[to]}; 44 else h[pos[to]].dis=dis[to]; 45 up(pos[to]); 46 } 47 pos[now]=0; 48 } 49 } 50 int main(){ 51 n=read(); 52 for(rg int i=1;i<=n;i++) 53 for(rg int j=i+1;j<=n;j++) e[++tot]=(edge){j,last[i],read()},last[i]=tot; 54 dijkstra(1); 55 printf("%d ",dis[n]); 56 return 0; 57 }