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

{

int
v,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)

{

int
vis[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
);

}

int main ()

{

int
n,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);

return
0;

}