SGU 438 - The Glorious Karlutka River =)(网络流‘最大流)

题目：

http://acm.sgu.ru/problem.php?contest=0&problem=438

题意：

河中有N块石头，河宽是W，M个游客，游客最多可以跳D米，每跳一次耗时1s。

给出N块石头的坐标和承重量，游客在南岸出发（即x轴）。问是否能够全部通过河，若能通过则求出最少用时。

思路：

动态流问题。在限制流量的情况下加入时间的限制。

通过枚举时间来建图，将每一时刻的流量加起来，得到的总流量超过人数则输出此刻的时间。

建立一个分层网络，随着时间的增加加入新的边，得到新的网络流量。对石头进行拆点。

AC.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;
int M, D, W;
struct node {
int x, y, c;
}dist[55];
bool d[55][55];

const int MAXN = 1e5+10;
const int MAXM = 8e5+10;
const int INF = 0x3f3f3f3f3f;
struct Edge
{
int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
edge[tol].to = v; edge[tol].cap = w; edge[tol].flow = 0;
edge[tol].next = head[u]; head[u] = tol++;
edge[tol].to = u; edge[tol].cap = rw; edge[tol].flow = 0;
edge[tol].next = head[v]; head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[end] = 0;
Q[rear++] = end;
while(front != rear)
{
int u = Q[front++];
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dep[v] != -1)continue;
Q[rear++] = v;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
}
}
}
int S[MAXN];
int sap(int start,int end, int N)
{
BFS(start,end);
memcpy(cur,head,sizeof(head));
int top = 0;
int u = start;
int ans = 0;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
int inser;

for(int i = 0;i < top;i++)
if(Min > edge[S[i]].cap - edge[S[i]].flow)
{
Min = edge[S[i]].cap - edge[S[i]].flow;
inser = i;
}
for(int i = 0;i < top;i++)
{
edge[S[i]].flow += Min;
edge[S[i]^1].flow -= Min;
}
ans += Min;
top = inser;
u = edge[S[top]^1].to;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = i;
break;
}
}
if(flag)
{
S[top++] = cur[u];
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1; i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if(u != start)u = edge[S[--top]^1].to;
}
return ans;
}
bool judge(int a, int b)
{
int dis = (dist[a].x - dist[b].x)*(dist[a].x - dist[b].x)
+ (dist[a].y - dist[b].y)*(dist[a].y - dist[b].y);
if(D*D >= dis) return true;
return false;
}

int main()
{
//freopen("in", "r", stdin);
int N;
while(~scanf("%d %d %d %d", &N, &M, &D, &W)) {
for(int i = 1; i <= N; ++i) {
scanf("%d%d%d", &dist[i].x, &dist[i].y, &dist[i].c);
}

if(D >= W) {
printf("1\n");
continue;
}

memset(d, 0, sizeof(d));
for(int i = 1; i <= N; ++i) {
for(int j = 1; j <= N; ++j) {
if(i != j && judge(i, j)) {
d[i][j] = 1;
}
}
}

int sum = 0;
int s = 0, t = (M+N+1)*50+5000+1, ok = 0;
init();

for(int ti = 1; ti <= N+M; ++ti) {
for(int i = 1; i <= N; ++i) {
addedge(ti*50+i, ti*50+5000+i, dist[i].c);

if(dist[i].y <= D) {
addedge(s, ti*50+i, INF);
}

if(dist[i].y + D >= W) {
addedge(ti*50+5000+i, t, INF);
}

for(int j = 1; j <= N; ++j) {
if(d[i][j]) {
addedge(ti*50+5000+i, (ti+1)*50+j, INF);
}
}
}
sum += sap(s, t, t+1);
//printf("%d\n", sum);
if(sum >= M) {
printf("%d\n", ti+1);
ok = 1;
break;
}
}
if(!ok) printf("IMPOSSIBLE\n");
}
return 0;
}