SGU 438 The Glorious Karlutka River =) ★(动态+分层网络流)
【题意】有一条东西向流淌的河,宽为W,河中有N块石头,每块石头的坐标(Xi, Yi)和最大承受人数Ci已知。现在有M个游客在河的南岸,他们想穿越这条河流,但是每个人每次最远只能跳D米,每跳一次耗时1秒。问他们能否全部穿越这条河流,如果能,最少需要多长时间。(0 <= N <= 50, 0 < M <= 50, 0 <= D <= 1000, 0 < W <= 1000, 0 < Xi < 1000, 0 < Yi < W, 0 <= Ci <= 1000)
非常好的题~分层图下的动态网络流~
【思路】
因为每个节点(石头)不同时间点的状态不同,所以按时间构造分层图是必须的。每块石头在每个时间点都有一个点表示。并且由于节点容量限制,需要拆点i, i'连一条容量限制边。然后所有能从南岸跳到的石头,从源点向其各时间点处连边,容量为∞,再建一个超级源点连向源点,容量为人数。所有能跳到北岸的石头,从其各时间点处向汇点连边,容量为∞。任意两块距离小于等于D的石头,互相从t到t+1连边,容量为∞。
接下来便是怎么求最小时间。一开始我的做法是二分验证,但是超时了。。。看到Edelweiss大牛《网络流建模汇总》中用的方法是动态网络流。就是枚举时间,不断地往网络中加点表示当前时刻的石头,直到最大流等于总人数为止。这里的“动态”当然就是动态加边的意思。然后我想了想也就明白了,这道题时间最大也就可能是N+M=100,二分的优势并不明显,相反,因为它需要不断重新构造图,所以耗费了很多的时间。而动态网络流此时效率就比较高了~
【总结】1.绿字部分 2.在遇到搜索答案+网络流验证的这种题目时,如果解区间较小,则用动态网络流效率高;如果解区间很大,便使用二分查找。
#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 10005;
const int MAXE = 2100005;
const int oo = 0x3fffffff;
struct node{
int u, v, flow;
int opp;
int next;
};
struct Dinic{
node arc[MAXE];
int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数
int cur[MAXV]; //当前弧
int q[MAXV]; //bfs建层次图时的队列
int path[MAXE], top; //存dfs当前最短路径的栈
int dep[MAXV]; //各节点层次
void init(int n){
vn = n;
en = 0;
mem(head, -1);
}
void insert_flow(int u, int v, int flow){
arc[en].u = u;
arc[en].v = v;
arc[en].flow = flow;
arc[en].opp = en + 1;
arc[en].next = head[u];
head[u] = en ++;
arc[en].u = v;
arc[en].v = u;
arc[en].flow = 0; //反向弧
arc[en].opp = en - 1;
arc[en].next = head[v];
head[v] = en ++;
}
bool bfs(int s, int t){
mem(dep, -1);
int lq = 0, rq = 1;
dep[s] = 0;
q[lq] = s;
while(lq < rq){
int u = q[lq ++];
if (u == t){
return true;
}
for (int i = head[u]; i != -1; i = arc[i].next){
int v = arc[i].v;
if (dep[v] == -1 && arc[i].flow > 0){
dep[v] = dep[u] + 1;
q[rq ++] = v;
}
}
}
return false;
}
int solve(int s, int t){
int maxflow = 0;
while(bfs(s, t)){
int i, j;
for (i = 1; i <= vn; i ++) cur[i] = head[i];
for (i = s, top = 0;;){
if (i == t){
int mink;
int minflow = 0x3fffffff;
for (int k = 0; k < top; k ++)
if (minflow > arc[path[k]].flow){
minflow = arc[path[k]].flow;
mink = k;
}
for (int k = 0; k < top; k ++)
arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
maxflow += minflow;
top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
i = arc[path[top]].u;
}
for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
int v = arc[j].v;
if (arc[j].flow && dep[v] == dep[i] + 1)
break;
}
if (j != -1){
path[top ++] = j;
i = arc[j].v;
}
else{
if (top == 0) break;
dep[i] = -1;
i = arc[path[-- top]].u;
}
}
}
return maxflow;
}
}dinic;
struct piles{
int x, y;
int cap;
}p[55];
vector reach[55];
vector rs, rt;
void check_reach(int n, int d, int w){
for (int i = 0; i <= n; i ++){
reach[i].clear();
}
rs.clear();
rt.clear();
for (int i = 1; i <= n; i ++){
if (p[i].y <= d)
rs.push_back(i);
if (p[i].y + d >= w)
rt.push_back(i);
for (int j = i + 1; j <= n; j ++){
if (sqrt((p[j].x - p[i].x)*(p[j].x - p[i].x) + (p[j].y - p[i].y)*(p[j].y - p[i].y)) <= d){
reach[i].push_back(j);
reach[j].push_back(i);
}
}
}
}
int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int n, m, d, w;
scanf("%d %d %d %d", &n, &m, &d, &w);
if (d >= w){
puts("1");
return 0;
}
for (int i = 1; i <= n; i ++){
scanf("%d %d %d", &p[i].x, &p[i].y, &p[i].cap);
}
check_reach(n, d, w);
int node_num = 2 * n * (n+m-1);
dinic.init(node_num+3);
dinic.insert_flow(node_num+3, node_num+1, m);
int time, res = 0;
for (time = 2; time <= n+m; time ++){
for (int i = 0; i < (int)rs.size(); i ++){
int stone = rs[i];
dinic.insert_flow(node_num+1, (2*stone-2)*(n+m-1)+time-1, oo);
}
for (int i = 0; i < (int)rt.size(); i ++){
int stone = rt[i];
dinic.insert_flow((2*stone-1)*(n+m-1)+time-1, node_num+2, oo);
}
for (int i = 1; i <= n; i ++){
dinic.insert_flow((2*i-2)*(n+m-1)+time-1, (2*i-1)*(n+m-1)+time-1, p[i].cap);
for (int j = 0; j < (int)reach[i].size(); j ++){
int stone = reach[i][j];
dinic.insert_flow((2*i-1)*(n+m-1)+time-1, (2*stone-2)*(n+m-1)+time, oo);
}
}
res += dinic.solve(node_num+3, node_num+2);
//printf("%d %d\n", time, res);
if (res == m){
break;
}
}
if (time > n+m){
puts("IMPOSSIBLE");
}
else{
printf("%d\n", time);
}
return 0;
}
举杯独醉,饮罢飞雪,茫然又一年岁。 ------AbandonZHANG