POJ 3169.Layout 最短路
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11612 | Accepted: 5550 |
Description
Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated.
Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.
Input
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.
Output
Sample Input
4 2 1
1 3 10
2 4 20
2 3 3
Sample Output
27
Hint
There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.
The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
const int maxn=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e13+;
struct edge
{
int from,to;
int cost;
};
int cou=;
edge es[maxn];
vector<edge>G[maxn];
int used[maxn];
priority_queue<P,vector<P>,greater<P> >que;
void addedge(int u,int v,int w)
{
cou++;
edge e;
e.from=u,e.to=v,e.cost=w;
es[cou].from=u,es[cou].to=v,es[cou].cost=w;
G[u].push_back(e);
}
int n,ml,md;
int al[maxn],bl[maxn],dl[maxn];
int ad[maxn],bd[maxn],dd[maxn];
int d[maxn];
void ford()
{
for(int i=; i<=n; i++) d[i]=inf;
d[]=;
for(int t=; t<n; t++)
{
for(int i=; i<n; i++)
if(d[i+]<inf) d[i]=min(d[i],d[i+]);
for(int i=; i<=ml; i++)
if(d[al[i]]<inf) d[bl[i]]=min(d[bl[i]],d[al[i]]+dl[i]);
for(int i=; i<=md; i++)
if(d[bd[i]]<inf) d[ad[i]]=min(d[ad[i]],d[bd[i]]-dd[i]);
}
if(d[]<) cout<<-<<endl;
else if(d[n]>=inf) cout<<-<<endl;
else cout<<d[n]<<endl;
}
int main()
{
int a,b,d;
scanf("%d%d%d",&n,&ml,&md);
for(int i=; i<n; i++) addedge(i+,i,);
for(int i=; i<=ml; i++)
scanf("%d%d%d",&al[i],&bl[i],&dl[i]);
for(int i=; i<=md; i++)
scanf("%d%d%d",&ad[i],&bd[i],&dd[i]);
ford();
return ;
}
/*
4 3 0
1 3 10
2 4 20
2 3 3
*/
最短路
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