差分约数:

求满足不等式条件的尽量小的值---->求最长路---->a-b>=c----> b->a (c)

Schedule Problem

3
2
3
4
SAF 2 1
FAF 3 2
#
3
1
1
1
SAF 2 1
SAF 3 2
SAF 1 3
#
0

Case 1:
1 0
2 2
3 1

Case 2:
impossible

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月29日 星期三 16时20分17秒
File Name     :HDOJ1534.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=5000;

int n;
int w[maxn];

struct Edge
{
	int to,next,cost;
}edge[maxn];

int Adj[maxn],Size;

void init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void Add_Edge(int u,int v,int c)
{
	edge[Size].to=v;
	edge[Size].cost=c;
	edge[Size].next=Adj[u];
	Adj[u]=Size++;
}

void SAF(int u,int v)
{
	Add_Edge(v,u,w[v]);
}

void SAS(int u,int v)
{
	Add_Edge(v,u,0);
}

void FAF(int u,int v)
{
	Add_Edge(v,u,w[v]-w[u]);
}

void FAS(int u,int v)
{
	Add_Edge(v,u,-w[u]);
}


/// spfa longest road

int dist[maxn],cq[maxn];
bool inq[maxn];

bool spfa()
{
	memset(dist,0xcf,sizeof(dist));
	memset(cq,0,sizeof(cq));
	memset(inq,false,sizeof(inq));

	dist[0]=0;
	queue<int> q;
	inq[0]=true; q.push(0);

	while(!q.empty())
	{
		int u=q.front(); q.pop();

		for(int i=Adj[u];~i;i=edge[i].next)
		{
			int v=edge[i].to;
			if(dist[v]<dist[u]+edge[i].cost)
			{
				dist[v]=dist[u]+edge[i].cost;
				if(!inq[v])
				{
					inq[v]=true;
					cq[v]++;
					if(cq[v]>=n) return false;
					q.push(v);
				}
			}
		}

		inq[u]=false;
	}

	return true; 
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int cas=1;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=1;i<=n;i++) scanf("%d",w+i);

		init();

		char cmd[20];
		while(scanf("%s",cmd)!=EOF)
		{
			if(cmd[0]=='#') break;

			int u,v;
			scanf("%d%d",&u,&v);

			if(strcmp(cmd,"SAF")==0)
			{
				SAF(u,v);
			}
			else if(strcmp(cmd,"FAF")==0)
			{
				FAF(u,v);
			}
			else if(strcmp(cmd,"FAS")==0)
			{
				FAS(u,v);
			}
			else if(strcmp(cmd,"SAS")==0)
			{
				SAS(u,v);
			}
		}

		for(int i=1;i<=n;i++) Add_Edge(0,i,0);
		bool fg=spfa();

		printf("Case %d:\n",cas++);

		if(fg==false)
		{
			puts("impossible");
		}
		else
		{
			int mx=0;
			for(int i=1;i<=n;i++)
			{
				if(dist[i]<mx) mx=dist[i];
			}
			for(int i=1;i<=n;i++)
			{
				printf("%d %d\n",i,dist[i]-mx);
			}
		}
		putchar(10);
	}
    
    return 0;
}

HDOJ 1534 Schedule Problem 差分约束

原文：http://blog.csdn.net/ck_boss/article/details/47130643