| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 28416 | Accepted: 10966 |
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.
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
Source
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
$S\left[ i\right] -S\left[ i-1\right] \geq 0$
$S\left[ i-1\right] -S\left[ i\right] \geq -1$
$S\left[ i-1\right] -S\left[ i\right] \geq -1$
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; }
