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 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.
第一次接觸查分約束還是有點懵的，利用了最短路的不等式dis[j]>dis[i]+len[i]來考慮不等式的問題，這里i和j要保持距離k以內，就往i到j構造一條長度為k的路徑，保持k距離以外，就往j到i構造一條權值為-k的路徑，出現負環即為不能構造出來（我決定最后應該再判斷每個點的位置才對）。
#include<iostream>#include<string.h>#include<stdio.h>#include<string>#include<vector>#include<algorithm>using namespace std;int n,ml,md;int from[20005], len[20005], to[20005],dis[1005];int enumber;bool is_ans;void add(int x, int y, int z){ to[enumber] = y; from[enumber] = x; len[enumber++] = z;}void bellman(){ for (int i = 1; i <= n; i++){ bool relax = 0; for (int j = 0; j < enumber; j++){ int a = from[j], b = to[j], c = len[j]; if (dis[b]>dis[a] + c){ relax = 1; dis[b] = dis[a] + c; } } if (relax == 0)break; if (relax&&i == n)is_ans = 0; }}int main(){ int x, y, z; while (scanf("%d%d%d", &n, &ml, &md) != EOF){ enumber = 0; for (int i = 0; i < ml; i++){ scanf("%d%d%d", &x, &y, &z); if (x > y)swap(x, y); add(x, y, z); } for (int i = 0; i < md; i++){ scanf("%d%d%d", &x, &y, &z); if (x > y)swap(x, y); add(y, x, -z); } memset(dis, 0x3f, sizeof(dis)); dis[1] = 0; is_ans = 1; bellman(); if (is_ans == 0){新聞熱點
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