BZoj 1003 物流运输 DP+最短路

2013-09-11 09:56

W[I]代表前I天能取得的最小花费，假设在第J天更改一次路线，那么如果有

W[I]>W[J]+第j+1到第I天的最小花费+更改路线的花费（K） 那么更新W[I]；

用最短路求第J+1到I的最短路*（I-J），边界则是W[1]=0;

因为最开始的路线不用更改（就是最初的路线不算在更改的费用中），这个

方程默认第一次的路线就是更改后的，多加了一次K，所以最后输出W[I]-K；

//By BLADEVIL
var
connect                 :array[0..110,0..110] of longint;
n, m, k, e, ch          :longint;
c                       :array[0..110,0..110] of longint;
w                       :array[0..110] of longint;
d                       :array[0..110] of longint;
vis                     :array[0..1100] of boolean;

procedure init;
var
i, j                    :longint;
x, y, z                 :longint;
begin
assign(input,'santajs.in'); reset(input);
assign(output,'santajs.out'); rewrite(output);
read(n,m,k,e);
fillchar(connect,sizeof(connect),10);
for i:=1 to e do
begin
read(x,y,z);
if z<connect[x,y] then
begin
connect[x,y]:=z;
connect[y,x]:=z;
end;
end;
read(ch);
for i:=1 to ch do
begin
read(x,y,z);
for j:=y to z do
begin
inc(c[j,0]);
c[j,c[j,0]]:=x;
end;
end;
end;

function dijkstra(s,t:longint):longint;
var
i, j, max, k            :longint;
begin
fillchar(vis,sizeof(vis),false);
for i:=1 to m do d[i]:=maxlongint;
d[1]:=0;
for i:=s to t do
for j:=1 to c[i,0] do
vis[c[i,j]]:=true;

for i:=1 to m-1 do
begin
max:=maxlongint;
for j:=1 to m do
if (not vis[j]) and (d[j]<max) then
begin
k:=j;
max:=d[j];
end;
vis[k]:=true;
for j:=1 to m do if d[j]>d[k]+connect[k,j] then d[j]:=d[k]+connect[k,j];
end;
exit(d[m]);
end;

procedure dp;
var
i, j, x                 :longint;
begin
for i:=1 to n do
begin
w[i]:=maxlongint div 10;
for j:=0 to i-1 do
begin
x:=dijkstra(j+1,i);
if x>100000000 then continue;
if w[i]>w[j]+k+(i-j)*x then w[i]:=w[j]+k+(i-j)*x;
end;
end;
writeln(w
-k);
close(input); close(output);
end;

begin
init;
dp;
end.