SGU 438 The Glorious Karlutka River =)(最大流)
Description
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
Output
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
- window安装ubuntu系统
- transform Vs Udf
在鞋厂的第一个任务,拆表.需要把订单表按照开始日期和结束日期拆分成多条记录,挺新鲜的~ transform方式,使用到了python. (1)把hive表的数据传入,通过python按照日期循环处理, ...
- python应用:日期时间
计算时间差时,注意天数差引发的问题,获取天数差为 (date2-date1).days 此处,需谨记date2>date1,以保证结果的正确性 具体应用如下: # -*-coding:utf8- ...
- python教程(二)·变量
什么是变量?在百度百科中,变量的解释是: 变量来源于数学,是计算机语言中能储存计算结果或能表示值抽象概念.变量可以通过-- 这是一段很长很长的解释,其实,作者认为没必要这么机械式的去理解.简单说,变量 ...
- Java学习笔记十五:Java中的成员变量和局部变量
Java中的成员变量和局部变量 一:成员变量: 成员变量在类中定义,用来描述对象将要有什么 成员变量可以被本类的方法使用,也可以被其他类的方法使用,成员变量的作用域在整个类内部都是可见的 二:局部变量 ...
- 【面试必问】python实例方法、类方法@classmethod、静态方法@staticmethod和属性方法@property区别
[面试必问]python实例方法.类方法@classmethod.静态方法@staticmethod和属性方法@property区别 1.#类方法@classmethod,只能访问类变量,不能访问实例 ...
- Java Web开发后端常用技术汇总
技术名称及官网 Spring Framework Spring容器 http://projects.spring.io/spring-framework/ SpringMVC Spring MVC框架 ...
- 北京Uber优步司机奖励政策(1月29日)
滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...
- Hadoop: Text类和String类的比较
一般认为Text类和String类是等价的,但二者之间其实存在着不小差别: 以<Hadoop权威指南>中的案例为例,给定字符串 String s = "\u0041\u00DF ...
- Ceres优化
Ceres Solver是谷歌2010就开始用于解决优化问题的C++库,2014年开源.在Google地图,Tango项目,以及著名的SLAM系统OKVIS和Cartographer的优化模块中均使用 ...