poj——3169Layout（差分约束 求第一只牛到第n只牛最大距离）

Layout

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10087 Accepted: 4853

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.

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.

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 11 3 102 4 202 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.

Source

USACO 2005 December Gold

/*分析：?对于任意i号奶牛，1<=i<N，在距离上应该满足：D[i+1] - D[i] >= 0对于每个好感的描述(i，j，k)，假设i<=j，体现到距离上的要求就是：D[j] - D[i] <= k对于每个反感的描述(i，j，k)，假设i<=j，体现到距离上的要求就是：D[j] - D[i] >= k写成我们约定的形式：D[i] - D[i+1] <= 0?D[j] -D[i ]<= kD[i] - D[j] <= - k1.对于差分不等式，a - b <= c ，建一条 b 到 a 的权值为 c 的边，求的是最短路，得到的是最大值（本题求的就是最大值），对于不等式 a - b >= c ，建一条 b 到 a 的权值为 c 的边，求的是最长路，得到的是最小值。2.如果检测到负环，那么无解。3.如果d[]没有更新，那么可以是任意解。*/ #include<iostream>#include<string>#include<cstring>#include<cstdio>using namespace std;int n,ml,md;const int INF=1<<24;const int maxnum=5000;int al[maxnum],bl[maxnum],dl[maxnum],ad[maxnum],bd[maxnum],dd[maxnum];int d[maxnum];void solve(){fill(d,d+n,INF);d[0]=0;for(int k=0;k<n;k++){for(int i=0;i+1<n;i++){if(d[i+1]<INF)d[i]=min(d[i],d[i+1]);}for(int i=0;i<ml;i++){if(d[al[i]-1]<INF)d[bl[i]-1]=min(d[al[i]-1]+dl[i],d[bl[i]-1]);}for(int i=0;i<md;i++){if(d[bd[i]-1]<INF){d[ad[i]-1]=min(d[ad[i]-1],d[bd[i]-1]-dd[i]);}}}}int main(){scanf("%d%d%d",&n,&ml,&md);for(int i=0;i<ml;i++){scanf("%d%d%d",&al[i],&bl[i],&dl[i]);}for(int i=0;i<md;i++){scanf("%d%d%d",&ad[i],&bd[i],&dd[i]);}solve();int res=d[n-1];if(d[0]<0)res=-1;else if(d[n-1]==INF)res=-2;printf("%d\n",res);return 0;}