【POJ 1716】Integer Intervals（差分约束系统）
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.
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
实训回来后的一血～～
差分约束系统，走前看了点，没搞透，做完这题略微有点明确了。
这题是差分约束系统入门题，关于差分约束系统。百度各种大牛博客讲的都非常具体。简单说就是通过不等关系建立约束系统图，然后跑最短路（大于关系则跑最长路）
回到此题，题目要求找出一个最小集合S，满足对于n个范围[ai,bi]，S中存在两个及两个以上不同的点在范围内
令Zi表示满足条件的情况下。0~i点至少有多少点在集合内
则Zb-Za >= 2
仅仅有这一个条件构造出来的图可能不是全然连通的，所以须要找一些“隐含条件”
不难发现 对于相邻的点 0 <= Zi-Z(i-1) <= 1 保证关系符同样 转化为
Zi-Z(i-1) >= 0
Z(i-1)-Zi >= -1
用这三个关系，就可以构造差分约束系统，然后SPFA或者Bellman跑一趟最长路（满足全部条件）
代码例如以下:
#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);
//最小点做起点 最大点做终点
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;
}