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POJ 2455 Secret Milking Machine

2012-08-16 18:48 204 查看
POJ_2455

每条路只走一次可以通过网络流来保证,而对于让最长的边最小可以通过二分枚举来搞定。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 210
#define MAXM 80010
#define INF 0x3f3f3f3f
int N, M, P, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM], MID;
int S, T, d[MAXD], q[MAXD], work[MAXD], tx[MAXM];
struct Edge
{
int x, y, z;
}edge[MAXM];
void add(int x, int y, int z)
{
v[e] = y, flow[e] = z;
next[e] = first[x], first[x] = e ++;
}
void init()
{
int i;
for(i = 0; i < M; i ++)
scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].z), tx[i] = edge[i].z;
std::sort(tx, tx + M);
}
int bfs()
{
int i, j, rear = 0;
memset(d, -1, sizeof(d[0]) * (N + 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;
for(int &i = work[cur]; i != -1; i = next[i])
if(flow[i] && d[v[i]] == d[cur] + 1)
if(int 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]) * (N + 1));
while(t = dfs(S, INF))
ans += t;
}
return ans;
}
void build()
{
int i;
S = 1, T = N;
memset(first, -1, sizeof(first[0]) * (N + 1)), e = 0;
for(i = 0; i < M; i ++)
if(edge[i].z <= MID)
add(edge[i].x, edge[i].y, 1), add(edge[i].y, edge[i].x, 1);
}
void solve()
{
int i, ans = 0, min = -1, max = M - 1, mid;
for(;;)
{
mid = (min + max + 1) >> 1;
MID = tx[mid];
if(mid == max) break;
build();
if(dinic() >= P)
max = mid;
else
min = mid;
}
printf("%d\n", MID);
}
int main()
{
while(scanf("%d%d%d", &N, &M, &P) == 3)
{
init();
solve();
}
return 0;
}
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