对差分约束理解还是不深入，约束条件总是漏掉。
题目中的约束条件
s[a]-s[b]>=n
s[i]-s[i-1]>=0
s[i]-s[i-1]<=1
s[i]表示整数集合Z从0到i这一范围内的整数个数
然后把第一个和第二个的约束乘以-1就可以编程最短了。
Intervals
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17050 | Accepted: 6383 |
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.
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.
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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define MAXN 50500
#define INF 0xFFFFFF
struct node
{
int to;
int dis;
};
vector<node> g[MAXN];
int dis[MAXN];
int l[MAXN],r[MAXN],num[MAXN];
bool in[MAXN];
int n,m,mm;
int mins,maxs;
void init()
{
node temp;
maxs=0;
mins=INF;
for(int i=1;i<=n;i++)
{
scanf("%d%d%d",&l[i],&r[i],&num[i]);
mins=min(l[i]-1,mins);
maxs=max(maxs,r[i]);
}
for(int i=mins;i<=maxs;i++)
g[i].clear();
for(int i=1;i<=n;i++)
{
temp.to=l[i]-1;
temp.dis=-num[i];
g[r[i]].push_back(temp);
}
for(int i=mins+1;i<=maxs;i++)
{
temp.to=i;
temp.dis=1;
g[i-1].push_back(temp);
temp.to=i-1;
temp.dis=0;
g[i].push_back(temp);
}
}
void spfa(int s)
{
queue<int> q;
for(int i=mins;i<=maxs;i++)
{
dis[i]=INF;
in[i]=false;
}
dis[s]=0;
in[s]=true;
q.push(s);
while(!q.empty())
{
int tag=q.front();
q.pop();
in[tag]=false;
for(int i=0;i<g[tag].size();i++)
{
int j=g[tag][i].to;
if(dis[j]>g[tag][i].dis+dis[tag])
{
dis[j]=g[tag][i].dis+dis[tag];
if(!in[j])
{
in[j]=true;
q.push(j);
}
}
}
}
}
int main()
{
while(cin>>n)
{
init();
spfa(maxs);
cout<<-dis[mins]<<endl;
}
return 0;
}