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SPOJ 287 Smart Network Administrator

2012-08-08 10:51 393 查看
SPOJ_287

YY了一下颜色的数量取决于网络流中各条边中最大的流量,于是二分边的容量并做网络流就可以了。

但至于为什么颜色的数量取决于网络流中各条边中最大的流量,暂时没有细加证明……

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 510
#define MAXM 501010
#define INF 0x3f3f3f3f
int N, M, K, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM], list[MAXD];
int S, T, d[MAXD], q[MAXD], work[MAXD];
struct Edge
{
int x, y;
}edge[MAXM];
void init()
{
int i;
scanf("%d%d%d", &N, &M, &K);
for(i = 0; i < K; i ++)
scanf("%d", &list[i]);
for(i = 0; i < M; i ++)
scanf("%d%d", &edge[i].x, &edge[i].y);
}
void add(int x, int y, int z)
{
v[e] = y, flow[e] = z;
next[e] = first[x], first[x] = e ++;
}
void build(int c)
{
int i;
memset(first, -1, sizeof(first[0]) * (T + 1));
e = 0;
add(S, 1, K), add(1, S, 0);
for(i = 0; i < K; i ++)
add(list[i], T, 1), add(T, list[i], 0);
for(i = 0; i < M; i ++)
add(edge[i].x, edge[i].y, c), add(edge[i].y, edge[i].x, c);
}
int bfs()
{
int i, j, rear = 0;
memset(d, -1, sizeof(d[0]) * (T + 1));
d[S] = 0, q[rear ++] = S;
for(i = 0; i < rear; i ++)
for(j = first[q[i]]; j != -1; j = next[j])
if(flow[j] && d[v[j]] == -1)
{
d[v[j]] = d[q[i]] + 1, q[rear ++] = v[j];
if(v[j] == T)
return 1;
}
return 0;
}
int dfs(int cur, int a)
{
if(cur == T)
return a;
int t;
for(int &i = work[cur]; i != -1; i = next[i])
if(flow[i] && d[v[i]] == d[cur] + 1)
if(t = dfs(v[i], std::min(a, flow[i])))
{
flow[i] -= t, flow[i ^ 1] += t;
return t;
}
return 0;
}
int dinic()
{
int ans = 0, t;
while(bfs())
{
memcpy(work, first, sizeof(first[0]) * (T + 1));
while(t = dfs(S, INF))
ans += t;
}
return ans;
}
void solve()
{
int min, max, mid;
S = 0, T = N + 1;
min = -1, max = K;
for(;;)
{
mid = min + max + 1 >> 1;
if(max == mid) break;
build(mid);
if(dinic() == K)
max = mid;
else
min = mid;
}
printf("%d\n", mid);
}
int main()
{
int t;
scanf("%d", &t);
while(t --)
{
init();
solve();
}
return 0;
}
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