入門題 put AD edge ota 全部 lib 最小 最短

id=1716">【POJ 1716】Integer Intervals（差分約束系統）

Integer Intervals

Time Limit: 1000MS	?	Memory Limit: 10000K
Total Submissions: 13425	?	Accepted: 5703

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b.
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.

Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

4
3 6
2 4
0 2
4 7

Sample Output

4

Source

field=source&key=CEOI+1997">CEOI 1997

實訓回來後的一血～～

差分約束系統，走前看了點，沒搞透，做完這題略微有點明確了。

這題是差分約束系統入門題，關於差分約束系統。百度各種大牛博客講的都非常具體。簡單說就是通過不等關系建立約束系統圖，然後跑最短路（大於關系則跑最長路）

回到此題，題目要求找出一個最小集合S，滿足對於n個範圍[ai,bi]，S中存在兩個及兩個以上不同的點在範圍內

令Zi表示滿足條件的情況下。0~i點至少有多少點在集合內

則Zb-Za >= 2

僅僅有這一個條件構造出來的圖可能不是全然連通的，所以須要找一些“隱含條件”

不難發現 對於相鄰的點 0 <= Zi-Z(i-1) <= 1 保證關系符同樣 轉化為

Zi-Z(i-1) >= 0

Z(i-1)-Zi >= -1

用這三個關系，就可以構造差分約束系統，然後SPFA或者Bellman跑一趟最長路（滿足全部條件）

代碼例如以下:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 1e5;
const int mod = 1e9+7;
const double eps = 1e-8;

struct Edge
{
	int v,w,next;
};

Edge eg[233333];
int head[50050];
bool vis[50050];
int dis[50050];
int tp,st,en;

void Add(int u,int v,int w)
{
	eg[tp].v = v;
	eg[tp].w = w;
	eg[tp].next = head[u];
	head[u] = tp++;
}

int SPFA()
{
	memset(vis,0,sizeof(vis));
	memset(dis,-INF,sizeof(dis));
	queue <int> q;
	dis[st] = 0;
	vis[st] = 1;
	int u,v,w;

	q.push(st);

	while(!q.empty())
	{
		u = q.front();
		q.pop();
		vis[u] = 0;
		for(int i = head[u]; i != -1; i = eg[i].next)
		{
			v = eg[i].v;
			w = eg[i].w;
			if(dis[v] < dis[u]+w)
			{
				dis[v] = dis[u]+w;
				if(!vis[v]) 
				{
					q.push(v);
					vis[v] = 1;
				}
			}
		}
	}
	return dis[en];
}

int main(int argc,char **argv)
{
	int n;
	int u,v;

	while(~scanf("%d",&n))
	{
		tp = 0;
		memset(head,-1,sizeof(head));
		
		en = 0,st = INF;

		while(n--)
		{
			scanf("%d%d",&u,&v);
			Add(u,v+1,2);
			//最小點做起點 最大點做終點
			en = max(en,v+1);
			st = min(st,u);
		}

		for(int i = st; i < en; ++i)
		{
			Add(i,i+1,0);
			Add(i+1,i,-1);
		}
		printf("%d\n",SPFA());
	}
	return 0;
}

【POJ 1716】Integer Intervals（差分約束系統）