POJ 1201 Intervals 差分约束

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23196		Accepted: 8762

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

解题思路

如果知道最终的串，那么就能求出到i时，所需要取的数的个数s[i]。

根据条件，可以知道，s[bi] - s[ai] >=ci;

将不等式变形后可得：s[bi] >= s[ai] + ci;

这个不等式不就是求最长路吗！！！

解题方法已经出来了-----差分约束

只要在mapt[ai][bi]建一条ci的边，跑最长路就行。

同时还有一些约束条件：

0<=s[i] - s[i-1] <=1

把这些条件都建好边，跑一边spfa，最后d
就是答案。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define maxn 50000
#define sf scanf
int n,m;
queue<int> que;
int d[maxn+10];
bool flag[maxn+10];
int p[maxn+10];
struct node{
int n,v,next;
}edge[maxn * 10];
void addedge(int x,int y,int k){edge[m].n = y;edge[m].v = k;edge[m].next = p[x];p[x] = m++;}
void spfa()
{
memset(d,-1,sizeof d);
memset(flag,true,sizeof flag);
while(!que.empty())que.pop();
que.push(0);
flag[0] = false;
d[0] = 0;
while(!que.empty())
{
int n = que.front();
que.pop();
// cout<<n<<endl;
int t = p
;
while(t!=-1)
{
int tmp = edge[t].n;
// cout<<"---"<<tmp<<endl;
if(d[tmp]<d
+edge[t].v)
{
d[tmp] = d
+ edge[t].v;
if(flag[tmp])
{
que.push(tmp);
flag[tmp] = false;
}
}
t = edge[t].next;
}
flag
= true;
}
}
int main()
{
while(sf("%d",&n)!=EOF)
{
memset(p,-1,sizeof p);
int m = 0;
for(int i =0;i<n;i++)
{
int x,y,z;
sf("%d%d%d",&x,&y,&z);
addedge(x-1,y,z);
}
for(int i = 0;i<maxn;i++)
{
addedge(i,i+1,0);
addedge(i+1,i,-1);
}
spfa();
//for(int i = 0;i<=11;i++)
//  cout<<i<<":"<<d[i]<<endl;
printf("%d\n",d[maxn]);
}
return 0;
}