【38.24%】【POJ 1201】Intervals

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 25902 Accepted: 9905

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

Source

Southwestern Europe 2002

【题目链接】:http://poj.org/problem?id=1201

【题解】



题意:

给你m个条件ai bi ci;

表示ai..bi这些整数在序列中至少出现了ci次；

然后问你这个序列最少能由多少个数字组成;

设d[i]表示1..i这些数字里面有多少个数字;

则所给的m个条件可以写出

d[bi]-d[ai]>=ci;

然后就转化为差分约束的题了；

考虑转化为最长路

if (d[u]-d[v]<c)
    d[u] = d[v]+c;

可以看到我们令d[u]=d[v]+c后实际上是让d[u]-d[v]=c;

而原本d[u]-d[v]是小于c的,而如果d[u]再变大一点也可以满足d[u]-d[v]>c但是显然直接让d[u]-d[v]=c会使得d[u]更小；

这样进行松弛操作后；整个d就是最小的了；

所以对输入的ai bi ci;

即d[bi]-d[ai]>=ci;

则在ai与bi之间建一条由ai指向bi的单向边,边权为ci;

又有0<=d[i]-d[i-1]<=1;->因为一个数字没必要出现两次；只可能有两种情况，出现一次或者不出现;

则

d[i]-d[i-1]>=0

d[i-1]-d[i]>=-1

再根据这两个在i-1和i之间建边权为0的边,在i和i-1之间建边权为-1的边;

然后一开始dis数组赋值为无穷小；然后从起点跑spfa即可;

但是这样还不够；有些细节要做到位

比如

2

1 2 1

2 3 1

这个输入

如果我们单纯地在ai与bi之间建一条由ai指向bi的单向边,边权为ci;

则会得到ans=2的错解;

原因在于程序没办法判断1-2和2-3是否重合了;

做法是把[ai,bi]换成[ai,bi+1);

因为是整数，所以这样的做法是正确的；

也即对于输入的ai bi ci；

直接令bi++;

然后再建边;

找到最左的端点作为s,最右的端点作为t;则从s开始跑spafa;最后输出d[t];



【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

const int MAXM = (50000+100)*4;
const int INF =-0x3f3f3f3f;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int w[MAXM],en[MAXM],fir[MAXM],nex[MAXM];
int n,totm=0;

void add(int x,int y,int z)
{
    totm++;
    nex[totm] = fir[x];
    fir[x] = totm;
    en[totm] = y;
    w[totm] = z;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    int s = 21e8,t=0;
    rei(n);
    rep1(i,1,n)
    {
        int x,y,z;
        rei(x);rei(y);rei(z);
        s = min(s,x);t=max(t,y+1);
        //dis[y]-dis[x]>=z
        add(x,y+1,z);
    }
    rep1(i,s,t)
    {
        add(i,i+1,0);
        add(i+1,i,-1);
    }
    int dis[MAXM];
    memset(dis,INF,sizeof(dis));
    dis[s] = 0;
    queue <int>dl;
    bool in[MAXM];
    dl.push(s);in[s] = true;
    while (!dl.empty())
    {
        int x = dl.front();
        in[x] = false;
        dl.pop();
        for (int temp = fir[x];temp;temp=nex[temp])
        {
            int y = en[temp];
            if (dis[y]<dis[x]+w[temp])
            {
                dis[y] = dis[x] + w[temp];
                if (!in[y])
                {
                    in[y] = true;
                    dl.push(y);
                }
            }
        }
    }
    cout << dis[t]<<endl;
    return 0;
}