POJ1201 Intervals(差分约束)
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 1Sample Output
6Source
题意：每次给出一段区间[a_i,b_i]以及一个数c_i，使得在这中间至少有c_i个数，求一个最小的集合Z，使得集合Z满足上述所有要求，问集合Z的大小
思路：
设S[i]表示0-i这一段区间的前缀和
那么题目的关系就变成了S[b_i]-S[a_i]>=c_i
这是一个很典型的差分约束类问题
题目中要求集合最小，因此转换为最长路，将所有的式子写成B-A>=C的形式
同时题目中还有一个条件0<=S[i]-S[i-1]<=1
因为数据为整数
于是又得到两个方程
S\left[ i\right] -S\left[ i-1\right] \geq 0 S\left[ i-1\right] -S\left[ i\right] \geq -1
但是有个细节：S[i-1]不能表示，因此我们需要将所有下标+1，此时S[i]表示0 to (i-1)的前缀和
同时，这个图一定是联通的，因此不用新建超级源点
#include<cstdio>
#include<queue>
#include<cstring>
#define INF 1e8+10
using namespace std;
const int MAXN=1e6+10;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,MAXN,stdin),p1==p2)?EOF:*p1++)
char buf[MAXN],*p1=buf,*p2=buf;
inline int read()
{
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
struct node
{
int u,v,w,nxt;
}edge[MAXN];
int head[MAXN],num=1;
int maxx=-INF,minn=INF;
int dis[MAXN],vis[MAXN];
inline void AddEdge(int x,int y,int z)
{
edge[num].u=x;
edge[num].v=y;
edge[num].w=z;
edge[num].nxt=head[x];
head[x]=num++;
}
int SPFA()
{
queue<int>q;
memset(dis,-0xf,sizeof(dis));
dis[minn]=0;q.push(minn);
while(q.size()!=0)
{
int p=q.front();q.pop();
vis[p]=0;
for(int i=head[p];i!=-1;i=edge[i].nxt)
{
if(dis[edge[i].v]<dis[p]+edge[i].w)
{
dis[edge[i].v]=dis[p]+edge[i].w;
if(vis[edge[i].v]==0)
vis[edge[i].v]=1,q.push(edge[i].v);
}
}
}
printf("%d",dis[maxx]);
}
int main()
{
#ifdef WIN32
freopen("a.in","r",stdin);
#else
#endif
memset(head,-1,sizeof(head));
int N=read();
for(int i=1;i<=N;i++)
{
int x=read(),y=read(),z=read();
AddEdge(x,y+1,z);
maxx=max(y+1,maxx);
minn=min(x,minn);
}
for(int i=minn;i<=maxx-1;i++)
{
AddEdge(i+1,i,-1);
AddEdge(i,i+1,0);
}
SPFA();
return 0;
}