poj 1201（差分约束）

Intervals

Description

You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn. 
Write a program that: 
reads the number of intervals, their end points and integers c1, ..., cn from the standard input, 
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n, 
writes the answer to the standard output. 

Input

The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.

Output

The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.

Sample Input

5 3 7 3 8 10 3 6 8 1 1 3 1 10 11 1

Sample Output

6

思路：

首先第一个转化，是找到一个合理的表示。用ti表示每一个数，如果有用就是1，否则是0。吧S(i+1)定义成S(i+1)=sigma(tj)(1<=j<=i)也就是。S[i+1]表示从0到i有多少个数是需要的。

因此，题目中的条件可以表示成S[bi+1]>=S[ai]+Ci//至少要Ci个

这与bellman中的松弛操作时很像的。因此可以看成一些点

有D[v]>=D[u]+w(u,v)

上式对任何u成立，所以v应该是里面最大的，若D[v]<D[u]+w(u,v)则D[v]=D[u]+w(u,v)

于是。可以从ai和bi+1连一条线，它的长度是ci

这里只有这些条件还是不够的，还要加上两个使其满足整数性质的条件

1>=s[i+1]-s[i]>=0

有了这么多条件，使其自然构成了一个差分约束系统。

用spfa算法得到一个最长路，第一个到最后一个节点的最长路即是需要求的值。

AC:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int maxn = 50005;
const int inf = 0x7f7f7f;
struct Edge
{
	int k,c,next;
}edge[maxn*10];
int n,pre[maxn],cnt,s,t,dis[maxn];
bool vis[maxn];

void addedge(int a,int b,int c)
{
	edge[cnt].k = b;
	edge[cnt].c = c;
	edge[cnt].next = pre[a];
	pre[a] = cnt++;
}

void SPFA()
{
	queue<int> que;
	memset(vis,false,sizeof(vis));
	for(int i = s; i <= t+1; i++)
		dis[i] = -inf;
	dis[s] = 0; vis[s] = true;
	que.push(s);
	while(!que.empty())
	{
		int u = que.front();
		que.pop();
		vis[u] = false;
		for(int i = pre[u]; i != -1; i = edge[i].next)
		{
			int k = edge[i].k;
			if(dis[k] < dis[u] + edge[i].c)
			{
				dis[k] = dis[u] + edge[i].c;
				if(vis[k] == false)
				{
					vis[k] = true;
					que.push(k);
				}
			}
		}
	}
	printf("%d\n",dis[t+1]);
}

int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		s = inf,t = -inf;
		memset(pre,-1,sizeof(pre));
		cnt = 0;
		for(int i = 1; i <= n; i++)
		{
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			s = min(a,s); t = max(b,t);
			addedge(a,b+1,c);
		}
		for(int i = s; i <= t; i++)
		{
			addedge(i,i+1,0);
			addedge(i+1,i,-1);
		}
		SPFA();
	}
	return 0;
}