hdu 6252 Subway Chasing（差分約束）
Subway Chasing
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 900 Accepted Submission(s): 299Special Judge
Problem Description
Mr. Panda and God Sheep are roommates and working in the same company. They always take subway to work together. There are N
Input
The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line consists of 3 integers N, M and X, indicating the number of stations, the number of chat contents and the minute interval between Mr. Panda and God Sheep. Then M lines follow, each consists of 4 integers A, B, C, D, indicating each chat content. 1≤T≤30 1≤N,M≤2000 1≤X≤109 1≤A,B,C,D≤NA≤B≤A+1C≤D≤C+1
Output
For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the minutes between stations in format t1,t2,...,tN−1. ti represents the minutes between station i and i+1. If there are multiple solutions, output a solution that meets 0<ti≤2×109. If there is no solution, output “IMPOSSIBLE” instead.
Sample Input
2 4 3 2 1 1 2 3 2 3 2 3 2 3 3 4 4 2 2 1 2 3 4 2 3 2 3
Sample Output
Case #1: 1 3 1 Case #2: IMPOSSIBLE
Hint
In the second test case, when God Sheep passed the third station, Mr. Panda hadn’t arrived the second station. They can not between the second station and the third station at the same time.
Source
Recommend
liuyiding | We have carefully selected several similar problems for you: 6447 6446 6445 6444 6443
題意：
就是給了兩個人的行走位置。b比a晚走x分鐘。然後兩個人在每一時刻彙報一下兩個人的位置。如果在確定點i，就彙報（i，i），如果在兩點中間，就彙報（i，i+1）.。。。問能否存在一個解使得彙報的成立，並輸出i~i+1之間的距離。
思路：
差分約束：
分為幾種情況：
1.當a,b,c,d都不相同時：
d-a<x && c-b>x
2.當a==b 時：
①當c==d時{
此時就是點到點的距離了，並且距離肯定是X。轉化為兩個不等關係：c-b>=x && c-b<=x
}
②當c！=d時，就是 d-a<x && d-c>x
3.當a！=b時：
①：當c==d時{
d-a<x && d-b>x
}
②：當c！=d時就轉化為 四個都不相同的情況了。
還有一點不要忘記：就是兩點之間的距離一定要大於等於1 ！！！！
根據上邊關係建邊即可。
建邊原則：
A-B<=C 就ADD(B,A,C)。。。
開始用的kuangbin大神的板子沒跑對。。。然後換了個板子就對了，還沒看出來kuangbin板子哪裡有問題。
程式碼：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;
#define LL long long
const int INF=0x3F3F3F3F;
int cnt;
int n,m;
LL x;
int u,uu,v,vv,vis[2009],tot[2009];
int s[2009],nt[200009],e[200009];
LL val[200009],dis[2009];
void add(int u,int v,int x){
nt[cnt]=s[u];
s[u]=cnt;
e[cnt]=v;
val[cnt++]=x;
}
int spfa()
{
memset(vis,0,sizeof vis);
memset(dis,0,sizeof dis);
memset(tot,0,sizeof tot);
queue<int>q;
q.push(1);
vis[1]=1;
tot[1]++;
while(!q.empty())
{
int pre=q.front();
q.pop();
vis[pre]=0;
for(int i=s[pre];~i;i=nt[i])
{
if(dis[e[i]]<dis[pre]+val[i])
{
dis[e[i]]=dis[pre]+val[i];
if(!vis[e[i]])
{
q.push(e[i]);
if((++tot[e[i]])>=n)
return 0;
vis[e[i]]=1;
}
}
}
}
return 1;
}
int main()
{
int t,a,b,c,d;
LL x;
int u,uu,v,vv;
int cas=0;
scanf("%d",&t);
while(t--){
cas++;
cnt=1;//加邊計數，這個不要忘
memset(s,-1,sizeof(s));
scanf("%d%d%lld",&n,&m,&x);
for(int i=1;i<=m;i++)
{
scanf("%d%d%d%d",&u,&uu,&v,&vv);
if(u==uu)
{
if(v==vv)
{
///就是兩點不一樣，中間距離肯定是x
add(u,v,x);///v-u<=x
add(v,u,-x);///u-v<=-x, v-u>=x
}
else
{
add(u,vv,x+1); /// vv-u < x
add(v,u,1-x); ///u-v<=1-x, v-u>x
}
}
else
{
if(v==vv)
{
add(u,v,x+1); ///v-u < x
add(v,uu,1-x); /// uu-v<=1-x v-uu> x
}
else
{
add(u,vv,x+1);///vv-u<=x+1 vv-u<x
add(v,uu,1-x);///uu-v>=1-x v-uu<x
}
}
}
for(int i=2;i<=n;i++)
add(i-1,i,1);
int flag=spfa();
printf("Case #%d:", cas);
if(!flag)
printf(" IMPOSSIBLE\n");
else
{
for(int i=2;i<=n;i++)
printf(" %lld",dis[i]-dis[i-1]);
printf("\n");
}
}
return 0;
}