poj 1201 Intervals
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Intervals
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 21863 Accepted: 8241
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
53 7 38 10 36 8 11 3 110 11 1
Sample Output
6
关键是怎么构造差分约束模型啊,这题设s[i]表示集合Z中小于等于i的个数,那么集合Z中[ai,bi]的个数就是s[bi]-s[ai-1],考虑到题目中ai最小为0,而s[ai-1]就越界了,因此将输入同步+1;
我们可以得到s[bi]-s[ai-1]>=ci,也就是s[ai-1]-s[bi]<=-ci,同时s[i]-s[i-1]<=1&&s[i-1]-s[i]<=0,之后就是简单的构造有向图了,我们来看看需要求的量,Z中元素的最小个数。
我们用变量maxx记录区间最右端的值,minx记录区间最左端的值,那么就是求|Z|=S[maxx]-S[minx-1]的最小值。假设S[maxx]-S[minx-1]>=M,则当M取最大时,左侧达到最小值。变型后S[minx-1]-S[maxx]<=-M,
M取大,也就是-M取小,也就是maxx->minx-1达到最短路径,M就是最短路径的相反数了。因此以maxx为源点,一遍最短路即可。
代码:
#include<cstdio>#include<iostream>#include<cstring>#define Maxn 50010using namespace std;struct line{ int v,w; line(int vv=0,int ww=0):v(vv),w(ww){}}p[Maxn*3];int head[Maxn],dist[Maxn],vis[Maxn],nxt[Maxn*3];int q[Maxn];int tot,l,r;const int inf=0x3f3f3f3f;void addedge(int u,int v,int w){ p[tot]=line(v,w); nxt[tot]=head[u]; head[u]=tot++;}void spfa(int u){ for(int i=l-1;i<=r;i++) dist[i]=inf,vis[i]=0; dist[u]=0; int s=0,e; q[e=1]=u; while(s!=e){ int v=q[s=(s+1)%Maxn]; vis[v]=0; for(int i=head[v];i!=-1;i=nxt[i]){ int r=p[i].v,w=p[i].w; if(dist[v]+w<dist[r]){ dist[r]=dist[v]+w; if(!vis[r]){ vis[r]=1; q[e=(e+1)%Maxn]=r; } } } }}int main(){ int n,fr,to,w; while(~scanf("%d",&n)){ memset(head,-1,sizeof head); tot=0; l=inf,r=0; for(int i=1;i<=n;i++){ scanf("%d%d%d",&fr,&to,&w); addedge(to+1,fr,-w); l=min(l,fr+1); r=max(r,to+1); } for(int i=l;i<=r;i++){ addedge(i-1,i,1); addedge(i,i-1,0); } spfa(r); printf("%d\n",-dist[l-1]); }return 0;}
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