POJ1201 Intervals【SPFA】【差分约束】

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writes the answer to the standard output. 

Output

Sample Input

Sample Output

Source

Southwestern Europe 2002

题目大意：给你N个整数点构成的区间[ai，bi](ai，bi都为整数)，在区间[ai，bi]上最少选ci个点。

ci可在区间[ai，bi]中随意取，但是不能重复。问：要满足在N个区间取点，至少要选多少个点。

思路：差分约束思想。设Si为前i项的整数个数，则S(bi) - S(ai-1) >= ci。还有两个隐含约束条件

S(i-1) - S(i) <= 0，S(i)-S(i-1) <= 1。把这三种约束构建一个差分约束系统，用SPFA求最短路径。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 50050;
const int INF = 0xffffff0;

struct EdgeNode
{
    int to;
    int w;
    int next;
}Edges[MAXN << 2];
int Head[MAXN],Dist[MAXN],vis[MAXN];

int SPFA(int Left,int Right)
{
    for(int i = Left; i <= Right+1; ++i)
        Dist[i] = INF;
    memset(vis,0,sizeof(vis));
    queue<int> Q;
    Q.push(Right);
    vis[Right] = 1;
    Dist[Right] = 0;
    while( !Q.empty() )
    {
        int u = Q.front();
        Q.pop();
        vis[u] = 0;
        for(int i = Head[u]; i != -1; i = Edges[i].next)
        {
            int temp = Dist[u] + Edges[i].w;
            if( temp < Dist[Edges[i].to])
            {
                Dist[Edges[i].to] = temp;
                if( !vis[Edges[i].to] )
                {
                    vis[Edges[i].to] = 1;
                    Q.push(Edges[i].to);
                }
            }
        }
    }
    return -Dist[Left];
}

int main()
{
    int N,Left,Right,a,b,c;
    while(~scanf("%d",&N))
    {
        Left = MAXN,Right = 0;
        memset(Head,-1,sizeof(Head));
        memset(Edges,0,sizeof(Edges));
        int id = 0;
        for(int i = 0; i < N; ++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            Edges[id].to = a;
            Edges[id].w = -c;
            Edges[id].next = Head[b+1];
            Head[b+1] = id++;
            if(a < Left)
                Left = a;
            if(b > Right)
                Right = b;
        }
        Right++;
        for(int i = Left; i < Right; ++i)
        {
            Edges[id].to = i;
            Edges[id].w = 0;
            Edges[id].next = Head[i+1];
            Head[i+1] = id++;

            Edges[id].to = i+1;
            Edges[id].w = 1;
            Edges[id].next = Head[i];
            Head[i] = id++;
        }

        printf("%d\n",SPFA(Left,Right));
    }

    return 0;
}

POJ1201 Intervals【SPFA】【差分约束】

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原文地址：http://blog.csdn.net/lianai911/article/details/43168263