POJ 1201 Intervals（差分约束+spfa 求最长路径）

题目链接：http://poj.org/problem?id=1201

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

题意：(转)

[ai, bi]区间内和点集Z至少有ci个共同元素，那也就是说如果我用Si表示区间[0,i]区间内至少有多少个元素的话，那么Sbi - Sai >= ci,这样我们就构造出来了一系列边，权值为ci，但是这远远不够，因为有很多点依然没有相连接起来（也就是从起点可能根本就还没有到终点的路线）,此时，我们再看看Si的定义，也不难写出0<=Si
- Si-1<=1的限制条件，虽然看上去是没有什么意义的条件，但是如果你也把它构造出一系列的边的话，这样从起点到终点的最短路也就顺理成章的出现了。

我们将上面的限制条件写为同意的形式：

Sbi - Sai >= ci

Si - Si-1 >= 0

Si-1 - Si >= -1

这样一来就构造出了三种权值的边，而最短路自然也就没问题了。

但要注意的是，由于查分约束系统里常常会有负权边，所以为了避免负权回路，往往用Bellman-Ford或是SPFA求解（存在负权回路则最短路不存在）。

PS:

因为求的是[ai,bi]区间，所以我们添加边的时候需要(u-1, v, w)!

把距离dis初始化为负无穷， if(dis[v] < dis[u] + w)即可！

代码如下：

#include <cstdio>
#include <cstring>
#include <stack>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define N 50017
#define M 50017
int n, m, k;
int Edgehead[N], dis[N];
struct Edge
{
    int v,w,next;
} Edge[3*M];
bool vis[N];
//int cont[N];
int minn, maxx;
int MIN(int a, int b)
{
    if(a < b)
        return a;
    return b;
}
int MAX(int a, int b)
{
    if(a > b)
        return a;
    return b;
}
void Addedge(int u, int v, int w)
{
    Edge[k].next = Edgehead[u];
    Edge[k].w = w;
    Edge[k].v = v;
    Edgehead[u] = k++;
}
int SPFA( int start)//stack
{
    int sta[N];
    int top = 0;
    for(int i = 1 ; i <= n ; i++ )
        dis[i] = -INF;
    dis[start] = 0;
    //++cont[start];
    memset(vis,false,sizeof(vis));
    sta[++top] = start;
    vis[start] = true;
    while(top)
    {
        int u = sta[top--];
        vis[u] = false;
        for(int i = Edgehead[u]; i != -1; i = Edge[i].next)//注意
        {
            int v = Edge[i].v;
            int w = Edge[i].w;
            if(dis[v] < dis[u] + w)
            {
                dis[v] = dis[u]+w;
                if( !vis[v] )//防止出现环
                {
                    sta[++top] = v;
                    vis[v] = true;
                }
                //	if(++cont[v] > n)//有负环
                //		return -1;
            }
        }
    }
    return dis[maxx];
}
int main()
{
    int u, v, w;
    while(~scanf("%d",&n))//n为目的地
    {
        k = 1;
        memset(Edgehead,-1,sizeof(Edgehead));
        minn = INF;
        maxx = -1;
        for(int i = 1 ; i <= n ; i++ )
        {
            scanf("%d%d%d",&u,&v,&w);
            Addedge(u-1,v,w);
            maxx = MAX(v,maxx);
            minn = MIN(u-1,minn);

        }
        for(int i = minn; i <= maxx; i++)//新边,保证图的连通性还必须添加每相邻两个整数点i,i+1的边
        {
            Addedge(i,i+1,0);
            Addedge(i+1,i,-1);
        }
        int ans = SPFA(minn);//从点minn开始寻找最短路
        printf("%d\n",ans);
    }
    return 0;
}

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