POJ 1201 Intervals 差分约束

Intervals

Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 23196
Accepted: 8762

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

53 7 38 10 36 8 11 3 110 11 1

Sample Output

6

解题思路

如果知道最终的串，那么就能求出到i时，所需要取的数的个数s[i]。

根据条件，可以知道，s[bi] - s[ai] >=ci;

将不等式变形后可得：s[bi] >= s[ai] + ci;

这个不等式不就是求最长路吗！！！

解题方法已经出来了-----差分约束

只要在mapt[ai][bi]建一条ci的边，跑最长路就行。

同时还有一些约束条件：

0<=s[i] - s[i-1] <=1

把这些条件都建好边，跑一边spfa，最后d[n]就是答案。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<queue>using namespace std;#define maxn 50000#define sf scanfint n,m;queue<int> que;int d[maxn+10];bool flag[maxn+10];int p[maxn+10];struct node{    int n,v,next;}edge[maxn * 10];void addedge(int x,int y,int k){edge[m].n = y;edge[m].v = k;edge[m].next = p[x];p[x] = m++;}void spfa(){    memset(d,-1,sizeof d);    memset(flag,true,sizeof flag);    while(!que.empty())que.pop();    que.push(0);    flag[0] = false;    d[0] = 0;    while(!que.empty())    {        int n = que.front();        que.pop();       // cout<<n<<endl;        int t = p[n];        while(t!=-1)        {            int tmp = edge[t].n;           // cout<<"---"<<tmp<<endl;            if(d[tmp]<d[n]+edge[t].v)            {                d[tmp] = d[n] + edge[t].v;                if(flag[tmp])                {                    que.push(tmp);                    flag[tmp] = false;                }            }            t = edge[t].next;        }        flag[n] = true;    }}int main(){    while(sf("%d",&n)!=EOF)    {        memset(p,-1,sizeof p);        int m = 0;        for(int i =0;i<n;i++)        {            int x,y,z;            sf("%d%d%d",&x,&y,&z);            addedge(x-1,y,z);        }        for(int i = 0;i<maxn;i++)        {            addedge(i,i+1,0);            addedge(i+1,i,-1);        }        spfa();        //for(int i = 0;i<=11;i++)          //  cout<<i<<":"<<d[i]<<endl;        printf("%d\n",d[maxn]);    }    return 0;}