这里的次短路是指最短路只有一条情况下的次短路,如果有两条及两条以上的最短路时则输出最短路 题目
求次短路的思路就是首先求出最短路后,记录最短路的路径,枚举删变(注意数组模拟链表要删双向边)后再次SPFA,求出即可。
program pro;
var
i,j,n,m,num,tot:longint;
min,cmin:real;
flag:boolean;
point:array[0..205]of record x,y:longint; end;
link:array[0..45000]of record ke,po,ne,br:longint; pr:real; bj:boolean; end;
pre:array[0..250]of record num,po:longint; end;
que:array[0..10000]of longint;
st:array[0..250]of longint;
dis:array[0..250]of real;
team:array[0..250]of boolean;
function fd(x,y:longint):real;
begin
exit(sqrt(sqr(point[x].x-point[y].x)+sqr(point[x].y-point[y].y)));
end;
procedure make(x,y:longint; z:real; br:longint);
begin
inc(tot);
link[tot].ne:=st[x];
st[x]:=tot;
link[tot].po:=y;
link[tot].ke:=x;
link[tot].pr:=z;
link[tot].br:=br;
end;
procedure init;
var
ii,u,v:longint;
c:real;
begin
readln(n,m);
for ii:=1 to n do readln(point[ii].x,point[ii].y);
for ii:=1 to m do
begin
readln(u,v);
c:=fd(u,v);
make(u,v,c,1);
make(v,u,c,-1);
end;
end;
procedure spfa(x:longint);
var
h,t,ke,po:longint;
temp:longint;
begin
fillchar(que,sizeof(que),0);
fillchar(dis,sizeof(dis),$5f);
fillchar(team,sizeof(team),false);
dis[x]:=0; team[x]:=true; que[1]:=x; h:=0; t:=1;
while h<=t do
begin
inc(h);
ke:=que[h]; team[ke]:=false;
temp:=st[ke];
while temp<>0 do
begin
po:=link[temp].po;
if (not link[temp].bj)and(dis[po]>dis[ke]+link[temp].pr) then
begin
dis[po]:=dis[ke]+link[temp].pr;
if not flag then
begin
pre[po].po:=ke;
pre[po].num:=temp;
end;
if (not team[po]) then
begin
inc(t);
que[t]:=po; team[po]:=true;
end;
end;
temp:=link[temp].ne;
end;
end;
flag:=true;
end;
procedure main;
var
po:longint;
begin
flag:=false;
fillchar(pre,sizeof(pre),0);
spfa(1);
min:=dis[n]; cmin:=maxlongint;
po:=n;
while po<>1 do
begin
num:=pre[po].num;
link[num].bj:=true;
link[num+link[num].br].bj:=true;
spfa(1);
if dis[n]=min then
begin
writeln(dis[n]:0:2);
closef;
halt;
end;
if (dis[n]>min)and(dis[n]<cmin) then cmin:=dis[n];
link[num].bj:=false;
link[num+link[num].br].bj:=false;
po:=pre[po].po;
end;
writeln(cmin:0:2);
end;
begin
init;
main;
end.