题目描述
Description
当排队等候喂食时,奶牛喜欢和它们的朋友站得靠近些。FJ有N(2<=N<=1000)头奶牛,编号从1到N,沿一条直线站着等候喂食。奶牛排在队伍中的顺序和它们的编号是相同的。因为奶牛相当苗条,所以可能有两头或者更多奶牛站在同一位置上。即使说,如果我们想象奶牛是站在一条数轴上的话,允许有两头或更多奶牛拥有相同的横坐标。
一些奶牛相互间存有好感,它们希望两者之间的距离不超过一个给定的数L。另一方面,一些奶牛相互间非常反感,它们希望两者间的距离不小于一个给定的数D。给出ML条关于两头奶牛间有好感的描述,再给出MD条关于两头奶牛间存有反感的描述。(1<=ML,MD<=10000,1<=L,D<=1000000)
你的工作是:如果不存在满足要求的方案,输出-1;如果1号奶牛和N号
奶牛间的距离可以任意大,输出-2;否则,计算出在满足所有要求的情况下,1号奶牛和N号奶牛间可能的最大距离。
输入描述 Input Description
Line 1: Three space-separated integers: N, ML, and MD.
Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.
Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.
输出描述 Output Description
Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.
样例输入 Sample Input
4 2 1
1 3 10
2 4 20
2 3 3
样例输出
Sample Output
27 首回学差分约束,一定要搞懂。//努力努力努力。。。。。。差分约束是一种通过图论的方法求最大值or最小值or最大差值or最小差值的方法 最大值:if dist[a]>dist[b]+map[a,b] then dist[a]:=dist[b]+map[a,b]; 即:dist[a]<=dist[b]+map[a,b] 即:dist[a]-dist[b]<=map[a,b] a-b<=c 就从b向a连一条权值为c的边 得到边,通过跑最短路
最小值:if dist[a]<dist[b]+map[a,b] then dist[a]:=dist[b]+map[a,b]; 即:dist[a]>=dist[b]+map[a,b] 即:dist[a]-dist[b]>=map[a,b] a-b>=c 就从b向a连一条权值为c的边 得到边,通过跑最长路 此题考察差分中边的转化,和最大差值题中有ml,md两种类型的边因为dist[1]=0已被设定,dist[n]值最大,差值最大转化为了求最大值最大值要求 a-b<=c 的边应该把md的边左右两边乘-1 变号还有,i-(i-1)>=0即:(i-1)-i<=0从i向(i-1)连一条权值为0的边再跑最短路。 code:
var n,ml,md:longint; x,y,z:longint; map:array[1..1000,1..1000]of longint; i,j,k:longint; mark:array[1..1000]of boolean; queue:array[1..20000]of longint; w:array[1..2000]of longint; head,tail:longint; renewal:array[1..1000]of longint;
function max(x,y:longint):longint; begin if x>y then exit(x) else exit(y); end;
function min(x,y:longint):longint; begin if x>y then exit(y) else exit(x); end;
begin readln(n,ml,md);
for i:=1 to n do for j:=1 to n do map[i,j]:=maxint*500; for i:=2 to n do map[i,i-1]:=0; for i:=1 to ml do begin readln(x,y,z); map[min(x,y),max(x,y)]:=z; end; for i:=1 to md do begin readln(x,y,z); map[max(x,y),min(x,y)]:=-1*z; end;
for i:=1 to n do w[i]:=maxint*500;
fillchar(mark,sizeof(mark),true); fillchar(queue,sizeof(queue),0); fillchar(renewal,sizeof(renewal),0); head:=1; tail:=n; for i:=1 to n do begin queue[i]:=i; mark[i]:=false; end; w[1]:=0;
while head<=tail do begin for i:=1 to n do begin if w[i]>w[queue[head]]+map[queue[head],i] then begin w[i]:=w[queue[head]]+map[queue[head],i]; inc(renewal[i]); if renewal[i]>n then begin writeln(‘-1‘); halt; end; if mark[i] then begin inc(tail); queue[tail]:=i; mark[i]:=false; end; end; end; mark[queue[head]]:=true; inc(head); end; if w[n]-w[1]=maxint*500 then writeln(‘-2‘) else writeln(w[n]-w[1]);end.