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): 4181 Accepted Submission(s): 1577


[align=left]Problem Description[/align]
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

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

[align=left]Sample Output[/align]

6

[align=left]Author[/align]
1384

一道很明显的差分约束题。
对于每个[l,r]对应的k可以列出方程 f[r]-f[l-1]>=k;
同时区间之间 f[i]-f[i-1]>=0; f[i-1]-f[i]>=-1;
然后就是拿所有区间的最左端点作源点跑一遍spfa（）;
结束。
代码：

1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<vector>
5 #include<set>
6 #define INF 1000000000
7 #define clr(x) memset(x,0,sizeof(x))
8 #define clrmin(x) memset(x,-1,sizeof(x))
9 #define clrmax(x) memset(x,0x3f3f3f3f,sizeof(x))
10 using namespace std;
11 struct node
12 {
13     int to,val,next;
14 }edge[50010*3];
15 int head[50010];
16 int dis[50010];
17 int n,maxx,minx,maxans,cnt,l,r,k;
18 set<int> Q;
19 void addedge(int l,int r,int k);
20 int min(int a,int b);
21 int max(int a,int b);
22 void spfa(int s);
23 int main()
24 {
25     while(scanf("%d",&n)!=EOF)
26     {
27         clrmin(head);
28         clrmin(dis);
29         Q.clear();
30         maxx=0;
31         cnt=0;
32         for(int i=1;i<=n;i++)
33         {
34             scanf("%d%d%d",&l,&r,&k);
35             addedge(l,r+1,k);
36             maxx=max(r+1,maxx);
37             minx=min(l,minx);
38         }
39         for(int i=minx;i<maxx;i++)
40         {
41             addedge(i,i+1,0);
42             addedge(i+1,i,-1);
43         };
44         spfa(minx);
45         printf("%d\n",dis[maxx]);
46     }
47     return 0;
48 }
49 void addedge(int l,int r,int k)
50 {
51     edge[++cnt].to=r;
52     edge[cnt].val=k;
53     edge[cnt].next=head[l];
54     head[l]=cnt;
55     return;
56 }
57 int min(int a,int b)
58 {
59     return a<b?a:b;
60 }
61 int max(int a,int b)
62 {
63     return a>b?a:b;
64 }
65 void spfa(int s)
66 {
67     dis[s]=0;
68     Q.insert(s);
69     int v,k;
70     while(!Q.empty())
71     {
72         v=*Q.begin();
73         Q.erase(v);
74         k=head[v];
75         while(k!=-1)
76         {
77             if(dis[v]+edge[k].val>dis[edge[k].to])
78             {
79                 dis[edge[k].to]=dis[v]+edge[k].val;
80                 Q.insert(edge[k].to);
81             }
82             k=edge[k].next;
83         }
84     }
85     return ;
86 }