Codeforces 543B Destroying Roads 【暴力 SPFA】

B. Destroying Roads

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

In some country there are exactly n cities and m bidirectional
roads connecting the cities. Cities are numbered with integers from 1 to n.
If cities a and b are
connected by a road, then in an hour you can go along this road either from city a to city b,
or from city b to city a.
The road network is such that from any city you can get to any other one by moving along the roads.

You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city s1 to
city t1 in
at most l1 hours
and get from city s2 to
city t2 in
at most l2 hours.

Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1.

Input

The first line contains two integers n, m (1 ≤ n ≤ 3000, 

) —
the number of cities and roads in the country, respectively.

Next m lines contain the descriptions of the roads as pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi).
It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them.

The last two lines contains three integers each, s1, t1, l1 and s2, t2, l2,
respectively (1 ≤ si, ti ≤ n, 0 ≤ li ≤ n).

Output

Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1.

Sample test(s)

input

5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 2

output

0

input

5 4
1 2
2 3
3 4
4 5
1 3 2
2 4 2

output

1

input

5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 1

output

-1

题意：给定n个点m条边的无向图(边权全为1)，让你去掉最多的边使得d(s1, t1) <= l1 && d(s2, t2) <= l2，若不能满足输出-1，反之输出可以去掉的最多边数。

思路：SPFA预处理所有点之间的距离。求出在满足d(s1, t1) <= l1 && d(s2, t2) <= l2的前提下，路径需要的最少边数ans，答案就是m - ans。

方法是：用dist[i][j]存储最短路。枚举d(s1, t1) 和 d(s2, t2)这两条路径上可能重合的路径d(i, j)

(1)D1=dist[s1][i] + dist[i][j] + dist[j][t1] <= l1 && D2=dist[s2][i] + dist[i][j] + dist[j][t2] <= l2

(2)D1=dist[s1][i] + dist[i][j] + dist[j][t1] <= l1 && D2=dist[s2][j] + dist[j][i] + dist[i][t2] <= l2

(3)D1=dist[s1][j] + dist[j][i] + dist[i][t1] <= l1 && D2=dist[s2][i] + dist[i][j] + dist[j][t2]
<= l2

(4)D1=dist[s1][j] + dist[j][i] + dist[i][t1]
<= l1 && D2=dist[s2][j] + dist[j][i] + dist[i][t2] <= l2

更新答案为ans = min(ans,  D1 + D2 - dist[i][j])。

AC代码；

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (3000+10)
#define MAXM (50000000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1#define rr o<<1|1using namespace std;
struct Edge{
int from, to, next;
};
Edge edge[MAXN*MAXN];
int head[MAXN], edgenum;
bool vis[MAXN];
int dist[MAXN][MAXN];
void init(){
edgenum = 0;
CLR(head, -1);
}
void addEdge(int u, int v)
{
Edge E = {u, v, head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
void SPFA(int s, int *d)
{
queue<int> Q;
CLR(vis, false);
d[s] = 0; vis[s] = true; Q.push(s);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next)
{
Edge E = edge[i];
if(d[E.to] > d[u] + 1)
{
d[E.to] = d[u] + 1;
if(!vis[E.to])
{
vis[E.to] = true;
Q.push(E.to);
}
}
}
}
}
int main()
{
int n, m;
Ri(n); Ri(m); init();
for(int i = 0; i < m; i++)
{
int a, b; Ri(a); Ri(b);
addEdge(a, b); addEdge(b, a);
}
int s1, t1, l1, s2, t2, l2;
Ri(s1); Ri(t1); Ri(l1);
Ri(s2); Ri(t2); Ri(l2);
CLR(dist, INF);
for(int i = 1; i <= n; i++)
SPFA(i, dist[i]);
if(dist[s1][t1] > l1 || dist[s2][t2] > l2)
printf("-1\n");
else
{
int ans = dist[s1][t1] + dist[s2][t2];
for(int i = 1; i <= n; i++)
{
for(int j = i+1; j <= n; j++)
{
if(dist[s1][i]+dist[i][j]+dist[j][t1] <= l1 && dist[s2][i]+dist[i][j]+dist[j][t2] <= l2)
ans = min(ans, dist[s1][i]+dist[i][j]+dist[j][t1]+dist[s2][i]+dist[j][t2]);
if(dist[s1][i]+dist[i][j]+dist[j][t1] <= l1 && dist[s2][j]+dist[j][i]+dist[i][t2] <= l2)
ans = min(ans, dist[s1][i]+dist[i][j]+dist[j][t1]+dist[s2][j]+dist[i][t2]);
if(dist[s1][j]+dist[j][i]+dist[i][t1] <= l1 && dist[s2][i]+dist[i][j]+dist[j][t2] <= l2)
ans = min(ans, dist[s1][j]+dist[j][i]+dist[i][t1]+dist[s2][i]+dist[j][t2]);
if(dist[s1][j]+dist[j][i]+dist[i][t1] <= l1 && dist[s2][j]+dist[j][i]+dist[i][t2] <= l2)
ans = min(ans, dist[s1][j]+dist[j][i]+dist[i][t1]+dist[s2][j]+dist[i][t2]);
}
}
Pi(m-ans);
}
return 0;
}