poj 1364 King 【非连通图的差分约束】【最长路 + 最短路求法】


King

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10878		Accepted: 3990

Description
Once, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: ``If my child was a son and if only he was a sound king.'' After nine months her child was born, and indeed, she gave birth to a nice
son.

Unfortunately, as it used to happen in royal families, the son was a little retarded. After many years of study he was able just to add integer numbers and to compare whether the result is greater or less than a given integer number. In addition, the numbers
had to be written in a sequence and he was able to sum just continuous subsequences of the sequence.

The old king was very unhappy of his son. But he was ready to make everything to enable his son to govern the kingdom after his death. With regards to his son's skills he decided that every problem the king had to decide about had to be presented in a form
of a finite sequence of integer numbers and the decision about it would be done by stating an integer constraint (i.e. an upper or lower limit) for the sum of that sequence. In this way there was at least some hope that his son would be able to make some decisions.

After the old king died, the young king began to reign. But very soon, a lot of people became very unsatisfied with his decisions and decided to dethrone him. They tried to do it by proving that his decisions were wrong.

Therefore some conspirators presented to the young king a set of problems that he had to decide about. The set of problems was in the form of subsequences Si = {aSi, aSi+1, ..., aSi+ni} of a sequence S = {a1, a2, ..., an}. The king thought a minute and then
decided, i.e. he set for the sum aSi + aSi+1 + ... + aSi+ni of each subsequence Si an integer constraint ki (i.e. aSi + aSi+1 + ... + aSi+ni < ki or aSi + aSi+1 + ... + aSi+ni > ki resp.) and declared these constraints as his decisions.

After a while he realized that some of his decisions were wrong. He could not revoke the declared constraints but trying to save himself he decided to fake the sequence that he was given. He ordered to his advisors to find such a sequence S that would satisfy
the constraints he set. Help the advisors of the king and write a program that decides whether such a sequence exists or not.

Input
The input consists of blocks of lines. Each block except the last corresponds to one set of problems and king's decisions about them. In the first line of the block there are integers n, and m where 0 < n <= 100 is length of the
sequence S and 0 < m <= 100 is the number of subsequences Si. Next m lines contain particular decisions coded in the form of quadruples si, ni, oi, ki, where oi represents operator > (coded as gt) or operator < (coded as lt) respectively. The symbols si, ni
and ki have the meaning described above. The last block consists of just one line containing 0.
Output
The output contains the lines corresponding to the blocks in the input. A line contains text successful conspiracy when such a sequence does not exist. Otherwise it contains text lamentable kingdom. There is no line in the output
corresponding to the last ``null'' block of the input.

Sample Input

4 2
1 2 gt 0
2 2 lt 2
1 2
1 0 gt 0
1 0 lt 0
0

Sample Output

lamentable kingdom
successful conspiracy

题意：一个长度为n的序列和它的m个子序列，每个子序列的和都有一个限制k，gt代表大于k，lt代表小于k。 问你这样的序列是否存在。

思路：大于k == 大于等于k+1, 小于k == 小于等于k-1。 图可能是不连通的，所以需要建立虚拟源点并和所有顶点连边来保证图的连通，接下来就是判断是否存在负环就可以了。

注意：图中有0到n共 n+1个点。

代码一：最短路

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 100+10
#define MAXM 10000+10
#define INF 1000000
using namespace std;
struct Edge
{
	int from, to, val, next;
}edge[MAXM];
int used[MAXN];//入队次数 
int dist[MAXN];//最短路
bool vis[MAXN];//标记是否入队
int head[MAXN], top;
int n, m;//序列长度 以及m个子序列 
void addEdge(int u, int v, int w)
{
	Edge E = {u, v, w, head[u]};
	edge[top] = E;
	head[u] = top++;
} 
void init()
{
	top = 0;
	memset(head, -1, sizeof(head));
	memset(vis, false, sizeof(vis));
	memset(used, 0, sizeof(used));
}
void getMap()
{
	int s, num, k;
	char op[3];
	while(m--)
	{
		scanf("%d%d%s%d", &s, &num, op, &k);
		switch(op[0])
		{
			case 'g': addEdge(s+num, s-1, -(k+1));break;//大于k 就是大于等于k+1 
			case 'l': addEdge(s-1, s+num, k-1);   break;//小于k 就是小于等于k-1 
		}
	}
	for(int i = 0; i <= n; i++)//连接虚拟源点 保证连通 
	addEdge(n+1, i, 0);
}
bool SPFA()
{
	queue<int> Q;
	for(int i = 0; i <= n+1; i++)
	dist[i] = i==n+1 ? 0 : INF; 
	Q.push(n+1);//n+1为虚拟源点 
	vis[n+1] = true;
	used[n+1]++; 
	while(!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		vis[u] = false;
		for(int i = head[u]; i != -1; i = edge[i].next) 
		{
			Edge E = edge[i];
			if(dist[E.to] > dist[u] + E.val)
			{
				dist[E.to] = dist[u] + E.val;
				if(!vis[E.to])
				{
					vis[E.to] = true;
					used[E.to]++;
					if(used[E.to] > n+1)//n+1个点 不算虚拟源点 
					return false;//存在负环 
					Q.push(E.to);
				} 
			}
		}
	}
	return true;//不存在负环 
} 
int main()
{
	while(scanf("%d", &n), n)
	{
		scanf("%d", &m);
		init();
		getMap();
		if(SPFA())
		printf("lamentable kingdom\n");//找到序列 
		else
		printf("successful conspiracy\n");
	}
	return 0;
}

看了下讨论区，又学会一个连通图的方法，直接把原话copy了下来。。。
差分约束根本不需要什么附加顶点, 附加顶点的唯一用处就是保证图的连通性, 不让你有负环判不到的情况,解决这种问题的最佳途径就是初始把所有顶点都加入队列, 并且将所有dis置0, 这就相当于加了一个不存在的附加顶点, 它与所有的顶点的直连长度都是0. 但是注意在判负环时必须是"cnt>n"而不是">=n"，因为第一次所有顶点入队只是相当于把一个附加顶点加入到队列中而已, 不应该算在cnt中, 如果在此步骤没有增加过cnt, 则">="也是可以的.

原来连通图 直接把所有顶点都先入队列就可以。

自己写了个代码：（AC）最短路

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 100+10
#define MAXM 10000+10
#define INF 1000000
using namespace std;
struct Edge
{
	int from, to, val, next;
}edge[MAXM];
int used[MAXN];//入队次数 
int dist[MAXN];//最短路
bool vis[MAXN];//标记是否入队
int head[MAXN], top;
int n, m;//序列长度 以及m个子序列 
void addEdge(int u, int v, int w)
{
	Edge E = {u, v, w, head[u]};
	edge[top] = E;
	head[u] = top++;
} 
void init()
{
	top = 0;
	memset(head, -1, sizeof(head));
	memset(used, 0, sizeof(used));
}
void getMap()
{
	int s, num, k;
	char op[3];
	while(m--)
	{
		scanf("%d%d%s%d", &s, &num, op, &k);
		switch(op[0])
		{
			case 'g': addEdge(s+num, s-1, -(k+1));break;//大于k 就是大于等于k+1 
			case 'l': addEdge(s-1, s+num, k-1);   break;//小于k 就是小于等于k-1 
		}
	}
}
bool SPFA()
{
	queue<int> Q;
	for(int i = 0; i <= n+1; i++)
	{
		dist[i] = 0; 
		Q.push(i);//全部顶点入队 
		vis[i] = true;//标记 
	} 
	while(!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		vis[u] = false;
		for(int i = head[u]; i != -1; i = edge[i].next) 
		{
			Edge E = edge[i];
			if(dist[E.to] > dist[u] + E.val)
			{
				dist[E.to] = dist[u] + E.val;
				if(!vis[E.to])
				{
					vis[E.to] = true;
					used[E.to]++;
					if(used[E.to] > n)
					return false;//存在负环 
					Q.push(E.to);
				} 
			}
		}
	}
	return true;//不存在负环 
} 
int main()
{
	while(scanf("%d", &n), n)
	{
		scanf("%d", &m);
		init();
		getMap();
		if(SPFA())
		printf("lamentable kingdom\n");//找到序列 
		else
		printf("successful conspiracy\n");
	}
	return 0;
}

最长路求法：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN 100+10
#define MAXM 100000
#define INF 1000000
using namespace std;
struct Edge
{
	int from, to, val, next;
}edge[MAXM];
int dist[MAXN];
int head[MAXN], top;
int used[MAXN];
bool vis[MAXN];
int n, m;
void init()
{
	top = 0;
	memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int w)
{
	Edge E = {u, v, w, head[u]};
	edge[top] = E;
	head[u] = top++;
}
void getMap()
{
	int a, b, c;
	char op[5];
	while(m--)
	{
		scanf("%d%d%s%d", &a, &b, op, &c);
		switch(op[0])
		{
			case 'g': addEdge(a-1, a+b, c+1); break;
			case 'l': addEdge(a+b, a-1, 1-c); break;
		}
	}
}
bool SPFA()
{
	queue<int> Q;
	for(int i = 0; i <= n; i++)
	{
		dist[i] = i==0 ? 0 : -INF;
		vis[i] = true;
		used[i] = 0;
		Q.push(i);//连通图 
	}
	while(!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		vis[u] = false;
		for(int i = head[u]; i != -1; i = edge[i].next)
		{
			Edge E = edge[i];
			if(dist[E.to] < dist[u] + E.val)
			{
				dist[E.to] = dist[u] + E.val;
				if(!vis[E.to])
				{
					vis[E.to] = true;
					used[E.to]++;
					if(used[E.to] > n) return false;
					Q.push(E.to);
				} 
			}
		}
	}
	return true;
}
int main()
{
	while(scanf("%d", &n), n)
	{
		scanf("%d", &m);
		init();
		getMap();
		if(SPFA())
		printf("lamentable kingdom\n");
		else
		printf("successful conspiracy\n");
	}
	return 0;
}