SGU 438 The Glorious Karlutka River =)（最大流）
Description
A group of Mtourists are walking along the Karlutka river. They want to cross the river, but they couldn't find a bridge. Fortunately, there are some piles of rubbish floating in the water, and the tourists have decided to try to cross the river by jumping from one pile to another.
A tourist can move up to D meters in any direction at one jump. One jump takes exactly one second. tourists know that the river is W meters wide, and they have estimated the coordinates of rubbish piles ( Xi, Yi) and the capacity of each pile ( Ci, the maximum number of tourists that this pile can hold at the same time). Rubbish piles are not very large and can be represented as points. The river flows along the X axis. tourists start on the river bank at 0 by Y axis. The Ycoordinate of the opposite bank is W.
tourists would like to know if they can get to the opposite bank of the river, and how long it will take.
A tourist can move up to D meters in any direction at one jump. One jump takes exactly one second. tourists know that the river is W meters wide, and they have estimated the coordinates of rubbish piles ( Xi, Yi) and the capacity of each pile ( Ci, the maximum number of tourists that this pile can hold at the same time). Rubbish piles are not very large and can be represented as points. The river flows along the X axis. tourists start on the river bank at 0 by Y axis. The Ycoordinate of the opposite bank is W.
tourists would like to know if they can get to the opposite bank of the river, and how long it will take.
Input
First line of input consists of four integers: number of rubbish piles N (0 ≤ N ≤ 50), number of tourists M (0 < M ≤ 50), maximum length of tourist's jump D (0 ≤ D ≤ 1000), and width of the river W (0 < W ≤ 1000) Following N lines describe the rubbish piles, each line consists of three integers: (0 < Xi < 1000, 0 < Yi< W, 0 ≤ Ci ≤ 1000) — pile coordinates and capacity.
Output
Output a single number indicating the minimal time (in seconds) in which all tourists will be able to cross the river, or the line " IMPOSSIBLE" if it is impossible to cross the river.
题目大意：M 个人要过河，,河宽为 W。河上有N个垃圾，给出垃圾的坐标以及垃圾能同时容纳的人数，现在这 M 个人一下能跳距离D，问最少需要多少时间才可以使所有人到达对岸。跳一步一秒。 （开始人都在 X 轴上.对岸可以看为 Y =W 的一条直线）
思路：二分答案，判定性的最大流。最少用时间1到达对岸，最后用时间N+M到达对岸（假设要经过全部垃圾并且全部垃圾都只能站一个人）。假设现在二分的答案为mid，那么对于每个垃圾分成mid-1份，每份代表一个时间，再每个时间分成两份P、P'。如果从X轴能走到某点P，那么对所有时间，S连一条边到P。如果从某点P能走到对岸，那么对所有时间，P'连一条边到T。然后对所有时间每个点P连一条边到P'，容量为该点能同时站多少人。最后所有距离在D以内的，对每个时间t，连边(P,t)→(Q,t+1)，(Q,t)→(P,t+1)。最大流即为用mid的时间能过去多少人。二分答案即可。这种做法正确是因为：每个点的时间t只能走到t+1的时间，而在每个时间里通过某一点的流量不大于该点的容量，在任何时刻都能从S出发和到达T。
另外
SAP（62MS）：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; const int MAXN = * * ;
const int MAXE = MAXN * ;
const int INF = 0x7fff7fff; struct SAP {
int vis[MAXN], head[MAXN];
int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
int to[MAXE], flow[MAXE], next[MAXE];
int ecnt, st, ed; void init(int ss, int tt) {
memset(head, , sizeof(head));
ecnt = ;
st = ss; ed = tt;
} void addEdge(int u,int v,int f) {
to[ecnt] = v; flow[ecnt] = f; next[ecnt] = head[u]; head[u] = ecnt++;
to[ecnt] = u; flow[ecnt] = ; next[ecnt] = head[v]; head[v] = ecnt++;
} int Max_flow() {
int ans = , minFlow = INF, n = ed, u;
for (int i = ; i <= n; ++i){
cur[i] = head[i];
gap[i] = dis[i] = ;
}
u = pre[st] = st;
gap[] = n;
while (dis[st] < n){
bool flag = false;
for (int &p = cur[u]; p; p = next[p]){
int v = to[p];
if (flow[p] > && dis[u] == dis[v] + ){
flag = true;
minFlow = min(minFlow, flow[p]);
pre[v] = u;
u = v;
if(u == ed){
ans += minFlow;
while (u != st){
u = pre[u];
flow[cur[u]] -= minFlow;
flow[cur[u] ^ ] += minFlow;
}
minFlow = INF;
}
break;
}
}
if (flag) continue;
int minDis = n-;
for (int p = head[u]; p; p = next[p]){
int v = to[p];
if (flow[p] && dis[v] < minDis){
minDis = dis[v];
cur[u] = p;
}
}
if (--gap[dis[u]] == ) break;
gap[dis[u] = minDis+]++;
u = pre[u];
}
return ans;
}
} G; const int MAX = ; struct Point {
int x, y, c;
}; int dis2(const Point &a, const Point &b) {
int xx = a.x - b.x, yy = a.y - b.y;
return xx * xx + yy * yy;
} int n, m, d, w;
Point rub[MAX]; #define FOR(i, s, t) for(int i = s; i <= t; ++i)
#define get_num(i, t) ((i - 1) * 2 * (mid - 1) + 2 * (t - 1) + 1) void solve() {
int left = , right = n + m + ;
while(left < right) {
int mid = (left + right) >> ;
int ss = n * (mid - ) * + , st = ss + , tt = ss + ;
G.init(st, tt);
G.addEdge(st, ss, m);
FOR(i, , n) {
if(rub[i].y <= d) {
FOR(t, , mid - ) G.addEdge(ss, get_num(i, t), INF);
}
if(rub[i].y + d >= w) {
FOR(t, , mid - ) G.addEdge(get_num(i, t) + , tt, INF);
}
}
FOR(i, , n) FOR(t, , mid - ) {
int num = get_num(i, t);
G.addEdge(num, num + , rub[i].c);
if(t < mid - ) G.addEdge(num + , num + , INF);
}
FOR(i, , n) FOR(j, , n) {
if(i == j || dis2(rub[i], rub[j]) > d * d) continue;
FOR(t, , mid - ) G.addEdge(get_num(i, t) + , get_num(j, t + ), INF);
}
int ans = G.Max_flow();
if(ans < m) left = mid + ;
else right = mid;
//printf("%d %d\n", mid, ans); system("pause");
}
if(right == n + m + ) puts("IMPOSSIBLE");
else printf("%d\n", right);
} int main() {
scanf("%d%d%d%d", &n, &m, &d, &w);
FOR(i, , n) scanf("%d%d%d", &rub[i].x, &rub[i].y, &rub[i].c);
FOR(i, , n) if(rub[i].c > m) rub[i].c = m;
if(w <= d) printf("1\n");
else solve();
}
DINIC（15MS）:
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; const int MAXN = * * ;
const int MAXE = MAXN * ;
const int INF = 0x7fff7fff; struct Dinic {
int n, m, st, ed, ecnt;
int vis[MAXN], head[MAXN];
int cur[MAXN], d[MAXN];
int to[MAXE], next[MAXE], flow[MAXE]; void init(){
memset(head,,sizeof(head));
ecnt = ;
} void addEdge(int u,int v,int f) {
to[ecnt] = v; flow[ecnt] = f; next[ecnt] = head[u]; head[u] = ecnt++;
to[ecnt] = u; flow[ecnt] = ; next[ecnt] = head[v]; head[v] = ecnt++;
} bool bfs() {
memset(vis, , sizeof(vis));
queue<int> que; que.push(st);
d[st] = ; vis[st] = true;
while(!que.empty()){
int u = que.front(); que.pop();
for(int p = head[u]; p; p = next[p]){
int v = to[p];
if (!vis[v] && flow[p] > ){
vis[v] = ;
d[v] = d[u] + ;
que.push(v);
if(v == ed) return true;
}
}
}
return vis[ed];
} int dfs(int u, int a) {
if(u == ed || a == ) return a;
int outflow = , f;
for(int &p = cur[u]; p; p = next[p]){
int v = to[p];
if(d[u] + == d[v] && (f = dfs(v, min(a, flow[p]))) > ){
flow[p] -= f;
flow[p ^ ] += f;
outflow += f;
a -= f;
if(a == ) break;
}
}
return outflow;
} int Maxflow(int ss, int tt, int nn) {
st = ss; ed = tt;
int ans = ;
while(bfs()){
for(int i = ; i <= ed; ++i) cur[i] = head[i];
ans += dfs(st, INF);
}
return ans;
}
} G; const int MAX = ; struct Point {
int x, y, c;
}; int dis2(const Point &a, const Point &b) {
int xx = a.x - b.x, yy = a.y - b.y;
return xx * xx + yy * yy;
} int n, m, d, w;
Point rub[MAX]; #define FOR(i, s, t) for(int i = s; i <= t; ++i)
#define get_num(i, t) ((i - 1) * 2 * (mid - 1) + 2 * (t - 1) + 1) void solve() {
int left = , right = n + m + ;
while(left < right) {
int mid = (left + right) >> ;
int ss = n * (mid - ) * + , st = ss + , tt = ss + ;
G.init();
G.addEdge(st, ss, m);
FOR(i, , n) {
if(rub[i].y <= d) {
FOR(t, , mid - ) G.addEdge(ss, get_num(i, t), INF);
}
if(rub[i].y + d >= w) {
FOR(t, , mid - ) G.addEdge(get_num(i, t) + , tt, INF);
}
}
FOR(i, , n) FOR(t, , mid - ) {
int num = get_num(i, t);
G.addEdge(num, num + , rub[i].c);
if(t < mid - ) G.addEdge(num + , num + , INF);
}
FOR(i, , n) FOR(j, , n) {
if(i == j || dis2(rub[i], rub[j]) > d * d) continue;
FOR(t, , mid - ) G.addEdge(get_num(i, t) + , get_num(j, t + ), INF);
}
int ans = G.Maxflow(st, tt, tt);
if(ans < m) left = mid + ;
else right = mid;
//printf("%d %d\n", mid, ans); system("pause");
}
if(right == n + m + ) puts("IMPOSSIBLE");
else printf("%d\n", right);
} int main() {
scanf("%d%d%d%d", &n, &m, &d, &w);
FOR(i, , n) scanf("%d%d%d", &rub[i].x, &rub[i].y, &rub[i].c);
FOR(i, , n) if(rub[i].c > m) rub[i].c = m;
if(w <= d) printf("1\n");
else solve();
}
BFS+ISAP（15MS）：
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; const int MAXN = * * ;
const int MAXE = MAXN * ;
const int INF = 0x7fff7fff; struct SAP {
int head[MAXN];
int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
int to[MAXE], flow[MAXE], next[MAXE];
int ecnt, st, ed, n; void init(int ss, int tt, int nn) {
memset(head, , sizeof(head));
ecnt = ;
st = ss; ed = tt; n = nn;
} void addEdge(int u,int v,int f) {
to[ecnt] = v; flow[ecnt] = f; next[ecnt] = head[u]; head[u] = ecnt++;
to[ecnt] = u; flow[ecnt] = ; next[ecnt] = head[v]; head[v] = ecnt++;
} void bfs() {
memset(dis, 0x3f, sizeof(dis));
queue<int> que; que.push(ed);
dis[ed] = ;
while(!que.empty()) {
int u = que.front(); que.pop();
++gap[dis[u]];
for(int p = head[u]; p; p = next[p]) {
int v = to[p];
if (dis[v] > n && flow[p ^ ] > ) {
dis[v] = dis[u] + ;
que.push(v);
}
}
}
} int Max_flow() {
int ans = , minFlow = INF, u;
for (int i = ; i <= n; ++i){
cur[i] = head[i];
gap[i] = dis[i] = ;
}
u = pre[st] = st;
//gap[0] = n;
bfs();
while (dis[st] < n){
bool flag = false;
for (int &p = cur[u]; p; p = next[p]){
int v = to[p];
if (flow[p] > && dis[u] == dis[v] + ){
flag = true;
minFlow = min(minFlow, flow[p]);
pre[v] = u;
u = v;
if(u == ed){
ans += minFlow;
while (u != st){
u = pre[u];
flow[cur[u]] -= minFlow;
flow[cur[u] ^ ] += minFlow;
}
minFlow = INF;
}
break;
}
}
if (flag) continue;
int minDis = n-;
for (int p = head[u]; p; p = next[p]){
int v = to[p];
if (flow[p] && dis[v] < minDis){
minDis = dis[v];
cur[u] = p;
}
}
if (--gap[dis[u]] == ) break;
gap[dis[u] = minDis+]++;
u = pre[u];
}
return ans;
}
} G; const int MAX = ; struct Point {
int x, y, c;
}; int dis2(const Point &a, const Point &b) {
int xx = a.x - b.x, yy = a.y - b.y;
return xx * xx + yy * yy;
} int n, m, d, w;
Point rub[MAX]; #define FOR(i, s, t) for(int i = s; i <= t; ++i)
#define get_num(i, t) ((i - 1) * 2 * (mid - 1) + 2 * (t - 1) + 1) void solve() {
int left = , right = n + m + ;
while(left < right) {
int mid = (left + right) >> ;
int ss = n * (mid - ) * + , st = ss + , tt = ss + ;
G.init(st, tt, tt);
G.addEdge(st, ss, m);
FOR(i, , n) {
if(rub[i].y <= d) {
FOR(t, , mid - ) G.addEdge(ss, get_num(i, t), INF);
}
if(rub[i].y + d >= w) {
FOR(t, , mid - ) G.addEdge(get_num(i, t) + , tt, INF);
}
}
FOR(i, , n) FOR(t, , mid - ) {
int num = get_num(i, t);
G.addEdge(num, num + , rub[i].c);
if(t < mid - ) G.addEdge(num + , num + , INF);
}
FOR(i, , n) FOR(j, , n) {
if(i == j || dis2(rub[i], rub[j]) > d * d) continue;
FOR(t, , mid - ) G.addEdge(get_num(i, t) + , get_num(j, t + ), INF);
}
int ans = G.Max_flow();
if(ans < m) left = mid + ;
else right = mid;
//printf("%d %d\n", mid, ans); system("pause");
}
if(right == n + m + ) puts("IMPOSSIBLE");
else printf("%d\n", right);
} int main() {
scanf("%d%d%d%d", &n, &m, &d, &w);
FOR(i, , n) scanf("%d%d%d", &rub[i].x, &rub[i].y, &rub[i].c);
FOR(i, , n) if(rub[i].c > m) rub[i].c = m;
if(w <= d) printf("1\n");
else solve();
}
SGU 438 The Glorious Karlutka River =)（最大流）的更多相关文章
- 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 ...
- SGU 0438 The Glorious Karlutka River =) 动态流
题目大意: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 ...
- SGU438 The Glorious Karlutka River =)（最大流）
题目大概说有m个人要过一条宽W的河,人最远跳远距离是d,河上有n个垃圾堆,每个垃圾堆都有坐标和同一时间能容纳的人数,问所有人最少要跳几次才能跳到对岸. 又是一题根据时间拆点的最大流. 二分时间建容量网 ...
- The Glorious Karlutka River =)
sgu438:http://acm.sgu.ru/problem.php?contest=0&problem=438 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( ...
- SGU438 The Glorious Karlutka River =)
传送门 sgu原来搬到cf了呀点了好几个链接才找到233 传说中的动态流(?) 反正很暴力就对了QwQ 有容量限制->拆点 对于每个点拆成入点和出点 时间限制->分层 对于每个时刻的每个石 ...
- SGU438_The Glorious Karlutka River =)
好题,有一些人在河的一边,想通过河里的某些点跳到对岸去.每个点最多只能承受一定数量的人,每人跳跃一次需要消耗一个时间.求所有人都过河的最短时间. 看网上说是用了什么动态流的神奇东东.其实就是最大流吧, ...
- SGU 185 Two shortest 最短路+最大流
题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=21068 Yesterday Vasya and Petya qua ...
- sgu 176 上下界网络流最小可行流带输出方案
算法步骤: 1. 先将原图像最大可行流那样变换,唯一不同的是不加dst->src那条边来将它变成无源无汇的网络流图.直接跑一边超级源到超级汇的最大流. 2. 加上刚才没有加上的那条边p 3. 再 ...
- Soj题目分类
-----------------------------最优化问题------------------------------------- ----------------------常规动态规划 ...
随机推荐
- 【Java】使用Atomic变量实现锁
Atomic原子操作 在 Java 5.0 提供了 java.util.concurrent(简称JUC)包,在此包中增加了在并发编程中很常用的工具类 Java从JDK1.5开始提供了java.uti ...
- 关于chrome浏览器不能更新js的问题
今天写程序时,突然发现无论我怎么改本地js,用chrome打开时,均是改动之前的效果,F12查看Sources时发现js文件并没有被改动.由此引发的问题,经查询解决方法如下: F12后按F1,出现Se ...
- python函数基本介绍
函数 1.函数结构 def 是函数的定义关键字,my_len是函数名.()传参用,冒号下面都是函数体. 执行函数方法:函数名加括号来执行函数.My_len() 举例: # s = 'lkfjsjulk ...
- Mysql 关于处理NULL值的相关函数和操作符
操作符 <=> NULL-safe equal. This operator performs an equality comparison like the = operator, bu ...
- ElasticSearch : APT-GET安装方式
Ubuntu下安装ES 新建非ROOT管理员用户 安装和配置JDK 打开下面的地址 https://www.elastic.co/guide/en/elasticsearch/reference/cu ...
- 大数据学习--day03(运算符、流程控制语句)
运算符.流程控制语句 自增自减容易出错的地方: 扩展的赋值运算符 a+=b 等同于 a = a+b; 扩展的赋值运算符 隐含了一个类型的强制转换 & && 有何区别 & ...
- 第3章 Hadoop 2.x分布式集群搭建
目录 3.1 配置各节点SSH无密钥登录 1.将各节点的秘钥加入到同一个授权文件中 2.拷贝授权文件到各个节点 3.测试无秘钥登录 3.2 搭建Hadoop集群 1.上传Hadoop并解压 2.配置H ...
- Zookeeper -- 本地\完全分布式 搭建
准备工作 linux软件:Zookeeper-3.4.12.tar.gz 四台centos系统虚拟机,主机名为:s101~s104 一.本地模式搭建(s101上安装) 1.解压软件压缩包:解压到根目录 ...
- linux 搭建ss
因为收藏的各种教程被xx,所以决定自己写 第一步.安装ss sudo pip install shadowsocks 第二步.配置IP.端口.密码.加密方式 vi /etc/shadowsocks.j ...
- php+IIS 配置环境（windows环境）
继php7+apache2.4 配置环境(window环境)后,由于B2C项目准备上线:特此小编在阿里云上搭建PHP7环境,为此特写上搭建过程希望正处于搭建php7+IIS(windows环境)中的朋 ...