HDU 1384 Intervals 差分约束

Intervals

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2756    Accepted Submission(s): 1005

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

53 7 38 10 36 8 11 3 110 11 1

 

Sample Output

6

 

Author

1384

 

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解题思路：差分约束，令si表示前i个数中选择了si个数，因为要防止i=0的情况发生，将区间[a,b]映射成[a+1,b+1]，还有隐含条件si+1-si>=1;si-si+1<=0;

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#define Maxn 50005#define Inf 0x3f3f3f3fusing namespace std;int head[Maxn],k;int dis[Maxn];struct{int e,w,next;}edge[Maxn<<2];void add(int s,int e,int w){edge[k].e=e;edge[k].w=w;edge[k].next=head[s];head[s]=k++;}int spfa(int s,int n){int i,st,ed;bool vis[Maxn];memset(vis,false,sizeof(vis));for(i=s;i<=n;i++)dis[i]=-Inf;dis[s]=0;vis[s]=true;queue<int> q;q.push(s);while(!q.empty()){st=q.front();q.pop();vis[st]=false;for(i=head[st];i!=-1;i=edge[i].next){ed=edge[i].e;if(dis[ed]<dis[st]+edge[i].w){dis[ed]=dis[st]+edge[i].w;if(!vis[ed]){vis[ed]=true;q.push(ed);}}}}return dis[n];}int main(){int n,a,b,c,maxn,ans,L,R;freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);while(~scanf("%d",&n)){k=0;ans=-1;memset(head,-1,sizeof(head));R=-Inf,L=Inf;while(n--){scanf("%d%d%d",&a,&b,&c);a++,b++;add(a-1,b,c);L=min(L,a-1);R=max(R,b);}for(int i=L;i<R;i++){add(i,i+1,0);add(i+1,i,-1);}printf("%d\n",spfa(L,R));}return 0;}