poj 1716 Integer Intervals（差分…

题意:给出N个整数区间[ai,bi],使得序列在区间[ai,bj]的个数>=2个,求出序列的最小长度
如样例：

3 6
 2 4
0 2
4 7
所对应的序列为：1 2 4 5

思路： dis[i] 表示 [0,i）中的元素个数，所以有 dis[v]-dis[u] >= 2还有个隐含条件 1>=dis[i+1]-dis[i]>=0
用spfa 实现，因为有负环所以用栈结构，队列是457MS 栈是32MS

852K   32MS

#include <stdio.h>
#include <string.h>
#define EM 40000
#define VM 10005
#define inf -10000000
struct edge
{
    intv,w,next;
}e[EM];
int head[VM],ep;

void addedge(int cu,int cv,int cw)
{
    ep ++;
    e[ep].v =cv;
    e[ep].w =cw;
    e[ep].next =head[cu];
    head[cu] =ep;
}
int maxn (int a,int b)
{
    return a> b?a:b;
}

void spfa (int n)
{
    intvis[VM],dis[VM],stack[EM];
    memset(vis,0,sizeof(vis));
    memset(dis,-1,sizeof(dis));
    dis[n+2] =0;    //超级源点
    int top =1;
    vis[n+2] =1;
    stack[0] =n+2;
    while(top)
    {
       int u = stack[--top];
       vis[u] = 0;
       for (int i = head[u];i != -1;i = e[i].next)
       {
           int v = e[i].v;
           if (dis[v] < dis[u] +e[i].w)   //松驰
           {
               dis[v] = dis[u] + e[i].w;
               if (!vis[v])
               {
                   vis[v] = 1;
                   stack[top++] = v;
               }
           }
       }
    }
    printf("%d\n",dis[n]);
}
int main ()
{
    intn,v1,v2,m;
    ep =0;
    scanf("%d",&n);
    m = n;
    memset(head,-1,sizeof(head));
    int max =-1;
    while (m--)
    {
       scanf ("%d%d",&v1,&v2);
       addedge (v1,v2+1,2);
       max = maxn (max,v2+1);
    }
    for (int i =0;i < max;i ++)
    {
       addedge (i,i+1,0);
       addedge (i+1,i,-1);
       addedge (max+2,i,0);  //定义一个超级源点，到其余点的距离为0
    }
    spfa(max);
    return0;
}