Intervals
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3332 Accepted Submission(s):
1227
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:给出n个区间的左右端点,和这个区间内至少存在的在集合s中的点的个数,让你求集合s中最少有多少个点
题解:重在找到差分约束的约束条件,将约束条件转化为xj-xi<=k的形式,然后建立一条从i到j权值为k的边;
设maxl为区间的左端点,maxr为区间的右端点,S[i] 表示集合Z里面的元素在区间[0, i ]的个数,Maxl,Maxr分别表示所有区间里面的最左端和最右端,dist[]数组存储源点到某点的最短路。则由题意得限制条件
一 S[right] - S[left-1] >= least 即[left, right]区间个数不小于least,转换得S[left-1] - S[right] <= least;
二 0 <= S[i] - S[i-1] <= 1转换得 S[i-1] - S[i] <= 0 && S[i] - S[i-1] <= 1。
第二个条件题中并没有给出,需要自己推导,因为仅仅靠题中的条件无法构建一个连通图,也就无法求最短路,因为s[i]表示的是集合Z里面的元素在区间[0, i ]的个数所以s[i]至多比s[i-1]大一也可能相等
然后根据限制条件建图
转化问题:题目需要求的是S[Maxr] - S[Maxl-1] >= ans 即S[Maxl-1] - S[Maxr] <= -ans。 若以Maxr为源点 ,而-ans就为Maxr到Maxl-1的最短路径的相反数,即-dist[Maxl-1]。
#include<stdio.h> #include<string.h> #include<queue> #define INF 0x3f3f3f #define MAX 200000 #include<algorithm> using namespace std; int n,ans; int maxl,maxr; int vis[MAX],dis[MAX]; int head[MAX]; struct node { int u,v,w; int next; }edge[MAX]; void add(int u,int v,int w) { edge[ans].u=u; edge[ans].v=v; edge[ans].w=w; edge[ans].next=head[u]; head[u]=ans++; } void init() { ans=0; maxl=INF; maxr=0; memset(head,-1,sizeof(head)); } void getmap() { int i,j,a,b,c; while(n--) { scanf("%d%d%d",&a,&b,&c); maxl=min(maxl,a); maxr=max(maxr,b); add(b,a-1,-c); } for(i=maxl;i<=maxr;i++) { add(i,i-1,0); add(i-1,i,1); } } void spfa() { int i,j; queue<int>q; memset(vis,0,sizeof(vis)); for(i=maxl-1;i<=maxr;i++)//以maxr为源点,也可以以maxl为源点,不过要对建图稍作修改 dis[i]=INF; dis[maxr]=0; vis[maxr]=1; q.push(maxr); while(!q.empty()) { int u=q.front(); q.pop(); vis[u]=0; for(i=head[u];i!=-1;i=edge[i].next) { int top=edge[i].v; if(dis[top]>dis[u]+edge[i].w) { dis[top]=dis[u]+edge[i].w; if(!vis[top]) { vis[top]=1; q.push(top); } } } } printf("%d\n",-dis[maxl-1]); } int main() { while(scanf("%d",&n)!=EOF) { init(); getmap(); spfa(); } return 0; }