POJ1201[Intervals]

Intervals
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 27184 Accepted: 10434
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

solution: 差分约束系统＋spfa

#include <cstdio>#include <iostream>#include <queue>#include <algorithm>#include <cstring>using namespace std;#define M 50010#define N M<<2int nxt[N], to[N], vl[N], head[M];int dist[N];bool vis[M];int cnt = 0;int MIN, MAX;void addeage( int u, int v, int w ){    nxt[++cnt] = head[u], to[cnt] = v, vl[cnt] = w, head[u] = cnt;}void SPFA(){    queue<int> q;    memset( dist, -0x3f3f3f3f, sizeof(dist));    memset( vis, false, sizeof(vis));    while ( !q.empty() ) q.pop();    q.push(MIN);    dist[MIN] = 0;    vis[MIN] = true;    while ( !q.empty() ){        int cur = q.front();        q.pop();        vis[cur] = false;        for ( int i = head[cur]; i; i = nxt[i] ){            int v = to[i];            if ( dist[v] < dist[cur] + vl[i] ){                dist[v] = dist[cur] + vl[i];                if ( !vis[v] ){                    vis[v] = true;                    q.push(v);                }            }        }    }}int main(){    int n;    scanf( "%d", &n );    for ( int i = 1; i <= n; i++ ){        int u, v, w;        scanf( "%d%d%d", &u, &v, &w );        addeage( u-1, v, w );        MIN = min( MIN, u-1);        MAX = max( MAX, v );    }    for ( int i = MIN; i <= MAX; i++){        addeage( i, i+1, 0);        addeage( i+1, i, -1);    }    SPFA();    printf( "%d", dist[MAX]);}