Intervals（poj1201）

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24134		Accepted: 9177

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. 

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
差分约束系统模板题；转换为最短路来求解，关键是建图；

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<algorithm>
#include<iostream>
#include<vector>
void spaf(int n,int ans);
using namespace std;
const int N=1e9;
int d[30005*5];
bool flag[30005*5];
void add(int x,int y,int z,int co);
typedef struct pp
{
    int x;
    int y;
    int cost;
    int pre;
}ss;ss aa[30005*5];
int id[30005*5];
int main(void) {
    int i,j,k,p,q;int z;
    int maxx=0;
    while(scanf("%d",&k)!=EOF) {int ans=0;
    fill(id,id+30005*5,-1);
        while(k--) {
            scanf("%d %d %d",&p,&q,&z);
            if(maxx<p)maxx=p;
            if(maxx<q)maxx=q;
            add(p-1,q,ans++,-z);
        }
        for(i=1; i<=maxx; i++) {
        add(i-1,i,ans++,0);
        add(i,i-1,ans++,1);
        }spaf(0,maxx);printf("%d\n",-d[maxx]);
    }
}

void spaf(int n,int ans) {
    fill(d,d+30005*5,N);
    d[n]=0;
    queue<int>que;int i;
    memset(flag,0,sizeof(flag));
    flag[n]=true;
    que.push(n);
    while(!que.empty()) {
        int c=que.front();
        que.pop();
        flag[c]=false;
        int x=id[c];
        while(x!=-1)
        {
            int uu=aa[x].y;
            if(d[uu]>d[c]+aa[x].cost)
            {d[uu]=d[c]+aa[x].cost;
            if(!flag[uu])
            {que.push(uu);
            flag[uu]=true;
            }
            }
            x=aa[x].pre;
        }

    }
}
void add(int x,int y,int z,int co)
{
    aa[z].x=x;
    aa[z].y=y;
    aa[z].cost=co;
    aa[z].pre=id[x];
    id[x]=z;
}

Integer Intervals

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13511		Accepted: 5756

Description

An integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. 
Write a program that: finds the minimal number of elements in a set containing at least two different integers from each interval.

Input

The first line of the input contains the number of intervals n, 1 <= n <= 10000. Each of the following n lines contains two integers a, b separated by a single space, 0 <= a < b <= 10000. They are the beginning and the end of an interval.

Output

Output the minimal number of elements in a set containing at least two different integers from each interval.

Sample Input

4
3 6
2 4
0 2
4 7

Sample Output

4

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #include<queue>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 void spaf(int n,int ans);
 9 using namespace std;
10 const int N=1e9;
11 int d[30005];
12 bool flag[30005];
13 void add(int x,int y,int z,int co);
14 typedef struct pp
15 {
16     int x;
17     int y;
18     int cost;
19     int pre;
20 }ss;ss aa[30005];
21 int id[30005];
22 int main(void) {
23     int i,j,k,p,q;
24     int maxx=0;
25     while(scanf("%d",&k)!=EOF) {int ans=0;
26     fill(id,id+30005,-1);
27         while(k--) {
28             scanf("%d %d",&p,&q);
29             p++;
30             q++;
31             if(maxx<p)maxx=p;
32             if(maxx<q)maxx=q;
33             add(p-1,q,ans++,-2);
34         }
35         for(i=1; i<=maxx; i++) {
36         add(i-1,i,ans++,0);
37         add(i,i-1,ans++,1);
38         }spaf(0,maxx);printf("%d\n",-d[maxx]);
39     }
40 }
41
42 void spaf(int n,int ans) {
43     fill(d,d+30005,N);
44     d[n]=0;
45     queue<int>que;int i;
46     memset(flag,0,sizeof(flag));
47     flag[n]=true;
48     que.push(n);
49     while(!que.empty()) {
50         int c=que.front();
51         que.pop();
52         flag[c]=false;
53         int x=id[c];
54         while(x!=-1)
55         {
56             int uu=aa[x].y;
57             if(d[uu]>d[c]+aa[x].cost)
58             {d[uu]=d[c]+aa[x].cost;
59             if(!flag[uu])
60             {que.push(uu);
61             flag[uu]=true;
62             }
63             }
64             x=aa[x].pre;
65         }
66
67     }
68 }
69 void add(int x,int y,int z,int co)
70 {
71     aa[z].x=x;
72     aa[z].y=y;
73     aa[z].cost=co;
74     aa[z].pre=id[x];
75     id[x]=z;
76 }