Layout
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5190 | Accepted: 2491 |
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：N头牛排队吃饭 排编号顺序排，大的永远在小的前面，但牛之间有的关系好，有的差，所以有的牛想离某些牛的距离最远不超过D 有的必须大于D 给出它们的关系 求第n头牛跟第一头的最远距离
思路：差分问题，求最大值 约束条件转化为 < ； 所以有S大-S小 <= D1,
S大-S小>=D2 把这个条件转化一下==> S小-S大<=-D2
然后就是spfa了
思路：差分问题，求最大值 约束条件转化为 < ； 所以有S大-S小 <= D1,
S大-S小>=D2 把这个条件转化一下==> S小-S大<=-D2
然后就是spfa了
#include <stdio.h> #include <string.h> #define VM 10005 #define EM 200500 //原先数据都是少一个0的，但RE 不知为什么 多加一个就过了 #define inf 9999999999 struct edge { int v,w,next; }e[EM]; int head[VM],ep; void addedge (int cu,int cv,int cw) { ep ++; e[ep].v = cv; e[ep].w = cw; e[ep].next = head[cu]; head[cu] = ep; } __int64 spfa (int n) { int vis[VM],cnt[VM],queue[VM]; __int64 dis[VM]; for (int i = 1;i <= n;i ++) { vis[i] = 0; cnt[i] = 0; dis[i] = inf; } dis[1] = 0; int front = -1,rear = 0; vis[1] = 1; cnt[1] = 1; queue[rear ++] = 1; while (front != rear) { front = (front+1)%(n+1); int u = queue[front]; vis[u] = 0; for (int i = head[u];i != -1;i = e[i].next) { int v = e[i].v; if (dis[v] > dis[u]+e[i].w) { dis[v] = dis[u] + e[i].w; if (!vis[v]) { vis[v] = 1; cnt[v] ++; if (cnt[v] >= n) return 0; //有负环输出-1 queue[rear] = v; rear = (rear+1)%(n+1); } } } } if (dis[n] == inf) return -1; //不可达点 else return dis[n]; } int main () { int n,ml,md; int v1,v2,cost; scanf ("%d%d%d",&n,&ml,&md); memset (head,-1,sizeof(head)); ep = 0; while (ml --) { scanf ("%d%d%d",&v1,&v2,&cost); if (v1 < v2) //编号大的牛永远在小的牛前面 addedge (v1,v2,cost); else addedge (v2,v1,cost); } while (md --) { scanf ("%d%d%d",&v1,&v2,&cost); if (v2 < v1) addedge (v1,v2,-1*cost); else addedge (v2,v1,-1*cost); } __int64 ans = spfa (n); if (ans == 0) printf ("-1\n"); else if (ans == -1) printf ("-2\n"); else printf ("%I64d\n",ans); return 0; }