POJ 3498 March of the Penguins

POJ_3498

对于任何一点来讲，限制的是跳走的企鹅的数量，如果这个点不是终点，那么必然跳过来的企鹅都会跳走，因此实际上限制的是经过这个点的企鹅的数量，这样通过拆点来限制经过点的企鹅的数量就可以了。最后暴力一点，枚举每个点作为终点并判断一下就可以了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 210
#define MAXM 40410
#define INF 0x3f3f3f3f
int N, first[MAXD], e, next[MAXM], v[MAXM], flow[MAXM];
int S, T, d[MAXD], q[MAXD], work[MAXD], list[MAXD], L, TOT;
double D;
struct Stone
{
double x, y;
int n, m;
}stone[MAXD];
double sqr(double x)
{
return x * x;
}
void init()
{
int i, j;
TOT = 0;
scanf("%d%lf", &N, &D);
for(i = 0; i < N; i ++)
scanf("%lf%lf%d%d", &stone[i].x, &stone[i].y, &stone[i].n, &stone[i].m), TOT += stone[i].n;
}
void add(int x, int y, int z)
{
v[e] = y, flow[e] = z;
next[e] = first[x], first[x] = e ++;
}
void build(int t)
{
int i, j;
S = 2 * N, T = t;
memset(first, -1, sizeof(first[0]) * (N << 1 | 1)), e = 0;
for(i = 0; i < N; i ++)
add(i, i + N, stone[i].m), add(i + N, i, 0);
for(i = 0; i < N; i ++)
for(j = i + 1; j < N; j ++)
if(sqr(D) >= sqr(stone[i].x - stone[j].x) + sqr(stone[i].y - stone[j].y))
add(i + N, j, INF), add(j, i + N, 0), add(j + N, i, INF), add(i, j + N, 0);
for(i = 0; i < N; i ++)
if(stone[i].n)
add(S, i, stone[i].n), add(i, S, 0);
}
int bfs()
{
int i, j, rear = 0;
memset(d, -1, sizeof(d[0]) * (N << 1 | 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 | 1));
while(t = dfs(S, INF))
ans += t;
}
return ans;
}
void solve()
{
int i;
L = 0;
for(i = 0; i < N; i ++)
{
build(i);
if(dinic() == TOT)
list[L ++] = i;
}
if(L == 0)
printf("-1\n");
else
{
printf("%d", list[0]);
for(i = 1; i < L; i ++) printf(" %d", list[i]);
printf("\n");
}
}
int main()
{
int t;
scanf("%d", &t);
while(t --)
{
init();
solve();
}
return 0;
}