hdu 1534 简单差分约束

Description

A project can be divided into several parts. Each part should be completed continuously. This means if a part should take 3 days, we should use a continuous 3 days do complete it. There are four types of constrains among these parts which are FAS, FAF, SAF and SAS. A constrain between parts is FAS if the first one should finish after the second one started. FAF is finish after finish. SAF is start after finish, and SAS is start after start. Assume there are enough people involved in the projects, which means we can do any number of parts concurrently. You are to write a program to give a schedule of a given project, which has the shortest time.

Input

The input file consists a sequences of projects.

Each project consists the following lines:

the count number of parts (one line) (0 for end of input)

times should be taken to complete these parts, each time occupies one line

a list of FAS, FAF, SAF or SAS and two part number indicates a constrain of the two parts

a line only contains a '#' indicates the end of a project

Output

Output should be a list of lines, each line includes a part number and the time it should start. Time should be a non-negative integer, and the start time of first part should be 0. If there is no answer for the problem, you should give a non-line output containing "impossible".

A blank line should appear following the output for each project.

Sample Input

3234SAF 2 1FAF 3 2#3111SAF 2 1SAF 3 2SAF 1 3#0

Sample Output

Case 1:1 02 23 1Case 2:impossible

应该是很明显的差分约束的题目了    只要想清楚   节点的含义   路径的含义   建好模  就好说

这里我直接设  d[I]  表示  第i个事件的开始时间

根据题意  可得

FAS：d[u]+val[u]>=d[v]

FAF：d[u]+val[u]>=d[v]+val[v]

SAF：d[u]>=d[v]+val[v]

SAS：d[u]>=d[v]

题目要求最小值，需要转化为：

d[u]--d[v]>=-val[u]

d[u]-d[v]>=val[v]-val[u]

d[u]-d[v]>=val[v]

d[u]-d[v]>=0

ac code

#include <iostream>#include <cstdio>#include <cmath>#include <queue>#include <cstring>#include <algorithm>#include <cstdlib>using namespace std;const int maxn=100005;const int inf=0x3f3f3f3f;int n,idx;int d[maxn],val[maxn],head[maxn],cnt[maxn];bool vis[maxn];struct node{    int v,w;    int nxt;}edge[maxn];void init(){    for(int i=0;i<=n;i++)    {        head[i]=-1;        d[i]=-inf;        vis[i]=false;        cnt[i]=0;    }    idx=0;}void add(int u,int v,int w){    edge[idx].v=v;    edge[idx].w=w;    edge[idx].nxt=head[u];    head[u]=idx++;}bool spfa(){    vis[0]=true;    d[0]=0;    queue<int> que;    que.push(0);    while(!que.empty())    {        int u=que.front();        que.pop();        vis[u]=false;        for(int id=head[u];id!=-1;id=edge[id].nxt)        {            int v=edge[id].v;            if(d[u]+edge[id].w>d[v])            {                d[v]=d[u]+edge[id].w;                if(!vis[v])                {                    vis[v]=true;                    que.push(v);                    if(++cnt[v]>n) return false;                }            }        }    }    return true;}int main(){    char ch[5];    int u,v,cas=0;    while(scanf("%d",&n),n)    {        init();        for(int i=1;i<=n;i++) scanf("%d",val+i);        scanf("%s",ch);        while(ch[0]!='#')        {            scanf("%d %d",&u,&v);            if(ch[0]=='S')            {                if(ch[2]=='S') add(v,u,0);                else add(v,u,val[v]);            }            else            {                if(ch[2]=='S')                    add(v,u,-val[u]);                else add(v,u,val[v]-val[u]);            }            scanf("%s",ch);        }        printf("Case %d:\n",++cas);        for(int i=1;i<=n;i++) add(0,i,0);        if(spfa()) for(int i=1;i<=n;i++) printf("%d %d\n",i,d[i]);        else puts("impossible");        puts("");    }    return 0;}