POJ 1201 Intervals

Intervals

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 1201
64-bit integer IO format: %lld      Java class name: Main

 

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

Source

Southwestern Europe 2002

 

解题：差分约束系统。参见书山有路，学海无涯

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 50010;
struct arc{
    int to,w,next;
    arc(int x = 0,int y = 0,int z = 0){
        to = x;
        w = y;
        next = z;
    }
};
int n,tot,head[maxn],theMin,theMax,d[maxn];
arc e[maxn<<2];
int q[maxn<<4],he,tail;
bool in[maxn];
void add(int u,int v,int w){
    e[tot] = arc(v,w,head[u]);
    head[u] = tot++;
}
int spfa(){
    for(int i = theMin; i <= theMax; i++){
        in[i] = false;
        d[i] = -INF;
    }
    d[theMin] = 0;
    he = tail = 0;
    q[tail++] = theMin;
    while(he < tail){
        int u = q[he++];
        in[u] = false;
        for(int i = head[u]; ~i; i = e[i].next){
            if(d[e[i].to] < d[u]+e[i].w){
                d[e[i].to] = d[u]+e[i].w;
                if(!in[e[i].to]){
                    in[e[i].to] = true;
                    q[tail++] = e[i].to;
                }
            }
        }
    }
    return d[theMax];
}
int main() {
    int u,v,w;
    while(~scanf("%d",&n)){
        memset(head,-1,sizeof(head));
        theMin = INF;
        theMax = -INF;
        for(int i = tot = 0; i < n; i++){
            scanf("%d %d %d",&u,&v,&w);
            add(u,v+1,w);
            theMin = min(theMin,u);
            theMax = max(theMax,v+1);
        }
        for(int i = theMin; i < theMax; i++){
            add(i,i+1,0);
            add(i+1,i,-1);
        }
        printf("%d\n",spfa());
    }
    return 0;
}