HDU 3592 World Exhibition

World ExhibitionTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission (s): 931 Accepted Submission (s): 468 Problem DescriptionNowadays, please people want to go to Shanghai to visit the World Exhibition. so there are always a lot of people who are standing along a straight line waiting for entering. assume that there are N (2 <= N <= 1,000) people numbered 1.. N who are standing in the same order as they are numbered. it is possible that two or more person line up at exactly the same location in the condition that those visit in a group. there is something interesting. some 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 X (1 <= X <= 10,000) constraints describes which person like each other and the maximum distance by which they may be separated; a subsequent list of Y constraints (1 <= Y <= 10,000) tells which person 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 person 1 and person N that satisfies the dista Nce constraints. inputFirst line: An integer T represents the case of test. the next line: Three space-separated integers: N, X, and Y. the next X lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A <B <= N. person A and B must be at most C (1 <= C <= 1,000,000) apart. the next Y lines: Each line contains three space-separated positive integers: A, B, and C, With 1 <= A <B <= C. person A and B must be at least C (1 <= C <= 1,000,000) apart. outputFor each line: A single integer. if no line-up is possible, output-1. if person 1 and N can be arbitrarily far apart, output-2. otherwise output the greatest possible distance between person 1 and N. sample Input14 2 11 3 82 4 152 3 4 Sample Output19 Authoralpc20 Source2010 ACM-ICPC Multi-University Trai Ning Contest (15) -- Host by NUDT Recommendzhouzeyong solution: the basic difference constraint, directly column the inequality, and then use SPFA to judge. If there is a negative weight ring, there is no inequality, if the shortest distance to the nth person is not updated, it means that the nth person and the ninth person are not connected on the graph, that is, the distance between the nth person and the ninth person is of any length, if there is an update, the shortest distance is reached, and the output is enough. [Cpp] # include <cstdio> # include <cstring> # include <queue> # define Inf 0x7ffffff using namespace std; int head [1005], k; struct {int s; int e; int w; int next;} edge [20005]; void Add (int s, int e, int w) {edge [k]. s = s; edge [k]. e = e; edge [k]. w = w; edge [k]. next = head [s]; head [s] = k ++;} int SPFA (int s) {int I, st, ed, dis [1005], cnt [1005]; bool visit [1005]; memset (visit, false, sizeof (visit); memset (cnt, 0, sizeof (Cnt); for (I = 1; I <= s; I ++) dis [I] = Inf; queue <int> x; x. push (1); dis [1] = 0; cnt [1] ++; visit [1] = true; while (! X. empty () {st = x. front (); x. pop (); for (I = head [st]; I! =-1; I = edge [I]. next) {ed = edge [I]. e; if (dis [ed]> dis [st] + edge [I]. w) {dis [ed] = dis [st] + edge [I]. w; if (! Visit [ed]) {visit [ed] = true; x. push (ed); cnt [ed] ++; if (cnt [ed]> s-1) return-1 ;}} visit [st] = false ;} if (dis [s] = Inf) return-2; else return dis [s];} int main () {int I, t, n, x, y,, b, c, ans; scanf ("% d", & t); while (t --) {k = 0; memset (head,-1, sizeof (head )); scanf ("% d", & n, & x, & y); for (I = 0; I <x; I ++) {scanf ("% d", & a, & B, & c); Add (a, B, c) ;}www.2cto.com for (I = 0; I <y; I ++) {scanf ("% d", & a, & B, & c); Add (B,, -c);} ans = SPFA (n); printf ("% d \ n", ans);} return 0 ;}