Layout
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12522 | Accepted: 6032 |
Description
Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).
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.
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
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.
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
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.
Sample Input
4 2 1 1 3 10 2 4 20 2 3 3
Sample Output
27
Hint
Explanation of the sample:
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.
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.
Source
题意：
我认为题意英文正常的都可以看的懂吧。
就是奶牛排队，一个地方可以容纳许多奶牛，
奶牛有互相喜欢的和互相讨厌的，
先输入喜欢的，1号3号相互喜欢，距离不能超过10
2和4不能超过20，2,3不能小于3
对于喜欢输入A,B,C
就是说d[B]-d[A]<=C;
转化为，d[B]<=C+d[A]；
求最多，用最短路，下面也需要化成形式一致的才可以。
不能忘了d[i]-d[i-1]>=0这个限制。
1 #include<cstring> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstdio> 6 #include<queue> 7 #define INF 2000000007 8 #define N 1007 9 #define M 10007 10 using namespace std; 11 12 int n,l,r; 13 int dis[N],num[N],ins[N]; 14 int cnt,head[N],next[M*3],rea[M*3],val[M*3]; 15 16 void add(int u,int v,int fee) 17 { 18 next[++cnt]=head[u]; 19 head[u]=cnt; 20 rea[cnt]=v; 21 val[cnt]=fee; 22 } 23 bool Spfa() 24 { 25 for (int i=1;i<=n;i++) 26 ins[i]=0,dis[i]=INF,num[i]=0; 27 queue<int>q; 28 q.push(1);dis[1]=0,num[1]=1; 29 while(!q.empty()) 30 { 31 int u=q.front();q.pop(); 32 for (int i=head[u];i!=-1;i=next[i]) 33 { 34 int v=rea[i],fee=val[i]; 35 if (dis[v]>dis[u]+fee) 36 { 37 dis[v]=dis[u]+fee; 38 if (!ins[v]) 39 { 40 num[v]++; 41 ins[v]=1; 42 q.push(v); 43 if (num[v]>n) return false; 44 } 45 } 46 } 47 ins[u]=0; 48 } 49 return true; 50 } 51 int main() 52 { 53 memset(head,-1,sizeof(head)); 54 scanf("%d%d%d",&n,&l,&r); 55 for (int i=1,x,y,z;i<=l;i++) 56 { 57 scanf("%d%d%d",&x,&y,&z); 58 add(x,y,z); 59 } 60 for (int i=1,x,y,z;i<=r;i++) 61 { 62 scanf("%d%d%d",&x,&y,&z); 63 add(y,x,-z); 64 } 65 for (int i=2;i<=n;i++) 66 add(i+1,i,0); 67 bool flag=Spfa(); 68 if (!flag) printf("-1 "); 69 else 70 { 71 if (dis[n]==INF) printf("-2 "); 72 else printf("%d ",dis[n]); 73 } 74 }