Problem 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 endpoints 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 <= 50 000) - 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 <= 50 000 and 1 <= ci <= bi - ai + 1.
Process to the end of file.
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
题意：对于每个区间[a,b]至少要有c个元素，问集合里元素的最小个数。
解析：差分约束，用Ti表示区间[0,i-1]有多少个元素在里面，则满足下面的条件
Tb+1-Ta>=c
0<=Ti+1-Ti<=1
则建边(a,b+1,c),(i,i+1,0),(i+1,i,-1) 然后用spfa求得答案。注意这题用vector可能会超时。
代码
#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
const int INF=1e9+7;
const double eps=1e-7;
const int maxn=50005;
int N;
struct edge
{
int u,v,w,next;
edge(int u=0,int v=0,int w=0):u(u),v(v),w(w){next=-1; }
}E[maxn*3];
int head[maxn],dist[maxn];
bool inq[maxn];
queue<int> que;
int spfa(int be,int en)
{
for(int i=be;i<=en;i++) dist[i]=-INF;
dist[be]=0;
memset(inq,false,sizeof(inq));
while(!que.empty()) que.pop();
que.push(be);
while(!que.empty())
{
int u=que.front(); que.pop();
inq[u]=false;
for(int i=head[u];i!=-1;i=E[i].next)
{
int v=E[i].v,w=E[i].w;
if(dist[v]<dist[u]+w) //更新
{
dist[v]=dist[u]+w;
if(!inq[v]){ inq[v]=true; que.push(v); }
}
}
}
return dist[en];
}
int main()
{
while(scanf("%d",&N)!=EOF)
{
memset(head,-1,sizeof(head));
int u,v,w,cnt=0;
int minv=INF,maxv=-INF;
for(int i=0;i<N;i++)
{
scanf("%d%d%d",&u,&v,&w); //建边(u,v+1,w);
v++;
E[++cnt]=edge(u,v,w);
E[cnt].next=head[u];
head[u]=cnt;
minv=min(minv,u);
maxv=max(maxv,v);
}
for(int i=minv;i<maxv;i++)
{
E[++cnt]=edge(i,i+1,0); //建边(i,i+1,0)
E[cnt].next=head[i];
head[i]=cnt;
E[++cnt]=edge(i+1,i,-1); //建边(i+1,i,-1)
E[cnt].next=head[i+1];
head[i+1]=cnt;
}
printf("%d\n",spfa(minv,maxv));
}
return 0;
}