FZU 1844 Earthquake Damage(最大流最小割)
Problem Description
Open Source Tools help earthquake researchers stay a step ahead. Many geological research facilities around the world use or are in the process of developing open source software and applications designed to interpret and share information with other researchers. For example, OpenSees is an open source software framework for developing apps that help understand what happens to structures during and after earthquakes to help engineers design stronger buildings. Researchers are also using OpenSees to understand the potential ill-effects of seismic activity on viaducts and bridges.

China has had an earthquake that has struck Sichuan Province on Monday 12 May 2008!
The earthquake has damaged some of the cities so that they are unpassable. Remarkably, after repairing by Chinese People's Liberation Army, all the driveways between cities were fixed.
As usual, Sichuan Province is modeled as a set of P (1 <= P <= 3,000) cities conveniently numbered 1..P which are connected by a set of C (1 <= C <= 20,000) non-directional driveways conveniently numbered 1..C. Driveway i connects city a_i and b_i (1 <= a_i <= P; 1 <= b_i <= P). Driveway might connect a_i to itself or perhaps might connect two cities more than once. The Crisis Center is located in city 1.
A total of N (1 <= N <= P) survivors (in different cities) sequentially contacts Crisis Center via moobile phone with an integer message report_j (2 <= report_j <= P) that indicates that city report_j is undamaged but that the calling survivor is unable to return to the Crisis Center from city report_j because he/she could not find a path that does not go through damaged city.
After all the survivors report in, determine the minimum number of cities that are damaged.
Input
Input consists of several testcases. The format of each case as follow:
- Line 1: Three space-separated integers: P, C, and N
- Lines 2..C+1: Line i+1 describes cowpath i with two integers: a_i and b_i
- Lines C+2..C+N+1: Line C+1+j contains a single integer: report_j
Output
For each testcase, output a line with one number, the minimum number of damaged cities.
用最大流最小割定理,最大流=最小割,很久以前做的,题目已忘,我是来放模版的
DINIC算法+当前弧优化+有容量上限,78MS
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std; #define MAXN 6010
#define MAXM 100010 #define INF 0x7fffffff struct Dinic {
int n, m, st, ed, ecnt;
int vis[MAXN], head[MAXN];
int cur[MAXN], d[MAXN];
int to[MAXM], next[MAXM], flow[MAXM], cap[MAXM]; void init(int ss, int tt){
memset(head,,sizeof(head));
ecnt = ;
st = ss; ed = tt;
} void addEdge(int u,int v,int c) {
//flow[ecnt] = c
to[ecnt] = v; cap[ecnt] = c; flow[ecnt] = ; next[ecnt] = head[u]; head[u] = ecnt++;
to[ecnt] = u; cap[ecnt] = ; 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] && cap[p] > flow[p]){//flow[p] > 0
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, cap[p] - flow[p]))) > ){//min(a, flow[p])
flow[p] += f;//flow[p] -= f;
flow[p ^ ] -= f;//flow[p ^ 1] += f;
outflow += f;
a -= f;
if(a == ) break;
}
}
return outflow;
} int Maxflow() {
int ans = ;
while(bfs()){
for(int i = ; i <= ed; ++i) cur[i] = head[i];
ans += dfs(st, INF);
}
return ans;
}
} G; int vis[MAXN]; int main() {
int ss, tt, N, C, P;
while(scanf("%d%d%d",&P,&C,&N)!=EOF){
ss = ; tt = *P+;
G.init(ss, tt);
while(C--){
int a, b;
scanf("%d%d",&a,&b);
G.addEdge(a + P, b, INF);
G.addEdge(b + P, a, INF);
}
memset(vis, , sizeof(vis));
while(N--){
int x;
scanf("%d",&x);
G.addEdge(x, tt, INF);
vis[x] = ;
}
for(int i = ; i <= P; ++i){
if(i != && !vis[i]) G.addEdge(i, i + P, );
else G.addEdge(i, i + P, INF);
}
G.n = tt;
printf("%d\n",G.Maxflow());
}
return ;
}
DINIC算法+当前弧优化+只有余量,62MS
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std; #define MAXN 6010
#define MAXM 100010 #define INF 0x7fffffff struct Dinic {
int n, m, st, ed, ecnt;
int vis[MAXN], head[MAXN];
int cur[MAXN], d[MAXN];
int to[MAXM], next[MAXM], flow[MAXM]; 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++;
} 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 ans = ;
while(bfs()){
for(int i = ; i <= ed; ++i) cur[i] = head[i];
ans += dfs(st, INF);
}
return ans;
}
} G; int vis[MAXN]; int main() {
int ss, tt, N, C, P;
while(scanf("%d%d%d",&P,&C,&N)!=EOF){
ss = ; tt = *P+;
G.init(ss, tt);
while(C--){
int a, b;
scanf("%d%d",&a,&b);
G.addEdge(a + P, b, INF);
G.addEdge(b + P, a, INF);
}
memset(vis, , sizeof(vis));
while(N--){
int x;
scanf("%d",&x);
G.addEdge(x, tt, INF);
vis[x] = ;
}
for(int i = ; i <= P; ++i){
if(i != && !vis[i]) G.addEdge(i, i + P, );
else G.addEdge(i, i + P, INF);
}
G.n = tt;
printf("%d\n",G.Maxflow());
}
return ;
}
ISAP算法,78MS(把注释删掉再交一次又变成了93MS,OJ的时间果然信不过o(╯□╰)o)//不能有点0
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std; #define MAXN 6010
#define MAXM 100010
#define INF 0x7fffffff struct SAP {
int vis[MAXN], head[MAXN];
int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
int to[MAXM], flow[MAXM], next[MAXM];
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; bool vis[MAXN]; int main() {
int ss, tt, N, C, P;
while(scanf("%d%d%d",&P,&C,&N)!=EOF) {
ss = ; tt = *P+;
G.init(ss, tt);
while(C--) {
int a, b;
scanf("%d%d",&a,&b);
G.addEdge(a + P, b, INF);
G.addEdge(b + P, a, INF);
}
memset(vis, , sizeof(vis));
while(N--) {
int x;
scanf("%d",&x);
G.addEdge(x, tt, INF);
vis[x] = ;
}
for(int i = ; i <= P; ++i) {
if(i != && !vis[i]) G.addEdge(i, i + P, );
else G.addEdge(i, i + P, INF);
}
printf("%d\n",G.Max_flow());
}
return ;
}
ISAP算法+BFS初始化,62MS(SGU 438把我这模板跑跪了,现在换掉了>_<)//不能有点0
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std; #define MAXN 6010
#define MAXM 100010
#define INF 0x7fffffff struct SAP {
int head[MAXN];
int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
int to[MAXM], flow[MAXM], next[MAXM];
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] > ed && 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; bool vis[MAXN]; int main() {
int ss, tt, N, C, P;
while(scanf("%d%d%d",&P,&C,&N)!=EOF) {
ss = ; tt = *P+;
G.init(ss, tt, tt);
while(C--) {
int a, b;
scanf("%d%d",&a,&b);
G.addEdge(a + P, b, INF);
G.addEdge(b + P, a, INF);
}
memset(vis, , sizeof(vis));
while(N--) {
int x;
scanf("%d",&x);
G.addEdge(x, tt, INF);
vis[x] = ;
}
for(int i = ; i <= P; ++i) {
if(i != && !vis[i]) G.addEdge(i, i + P, );
else G.addEdge(i, i + P, INF);
}
printf("%d\n",G.Max_flow());
}
return ;
}
FZU 1844 Earthquake Damage(最大流最小割)的更多相关文章
- bzoj 1585: [Usaco2009 Mar]Earthquake Damage 2 地震伤害
1585: [Usaco2009 Mar]Earthquake Damage 2 地震伤害 Description Farmer John的农场里有P个牧场,有C条无向道路连接着他们,第i条道路连接着 ...
- hiho 第116周,最大流最小割定理,求最小割集S,T
小Hi:在上一周的Hiho一下中我们初步讲解了网络流的概念以及常规解法,小Ho你还记得内容么? 小Ho:我记得!网络流就是给定了一张图G=(V,E),以及源点s和汇点t.每一条边e(u,v)具有容量c ...
- hihocoder 网络流二·最大流最小割定理
网络流二·最大流最小割定理 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 小Hi:在上一周的Hiho一下中我们初步讲解了网络流的概念以及常规解法,小Ho你还记得内容么? ...
- [HihoCoder1378]网络流二·最大流最小割定理
思路: 根据最大流最小割定理可得最大流与最小割相等,所以可以先跑一遍EdmondsKarp算法.接下来要求的是经过最小割切割后的图中$S$所属的点集.本来的思路是用并查集处理所有前向边构成的残量网络, ...
- HDU 1569 方格取数(2)(最大流最小割の最大权独立集)
Description 给你一个m*n的格子的棋盘,每个格子里面有一个非负数. 从中取出若干个数,使得任意的两个数所在的格子没有公共边,就是说所取数所在的2个格子不能相邻,并且取出的数的和最大. ...
- 【codevs1907】方格取数3(最大流最小割定理)
网址:http://codevs.cn/problem/1907/ 题意:在一个矩阵里选不相邻的若干个数,使这些数的和最大. 我们可以把它看成一个最小割,答案就是矩阵中的所有数-最小割.先把矩阵按国际 ...
- 洛谷 P2932 [USACO09JAN]地震造成的破坏Earthquake Damage
P2932 [USACO09JAN]地震造成的破坏Earthquake Damage 题目描述 Wisconsin has had an earthquake that has struck Farm ...
- 紫书 例题 11-12 UVa 1515 (最大流最小割)
这道题要分隔草和洞, 然后刘汝佳就想到了"割"(不知道他怎么想的, 反正我没想到) 然后就按照这个思路走, 网络流建模然后求最大流最小割. 分成两部分, S和草连, 洞和T连 外围 ...
- HDU-4289-Control(最大流最小割,拆点)
链接: https://vjudge.net/problem/HDU-4289 题意: You, the head of Department of Security, recently receiv ...
随机推荐
- RAC初体验(环境搭建)
实施阶段: 1.主机配置 2.安装Clusterware 3.安装Oracle Database 4.配置Listener 5.创建ASM 6.创建Database 一.主机配置 1.网络设置 I ...
- Python对文件目录的操作
python中对文件.文件夹(文件操作函数)的操作需要涉及到os模块和shutil模块. 得到当前工作目录,即当前Python脚本工作的目录路径: os.getcwd()返回指定目录下的所有文件和目录 ...
- php源码建博客1--搭建站点-实现登录页面
主要: 站点搭建 实现登录页面 分析及改进 站点搭建 1) 在apache安装目录下: [conf\extra\httpd-vhosts.conf]加入站点配置 <VirtualHost *: ...
- 使用TryParse()来执行数值转换
static void Main() { var ageText = "25"; if (int.TryParse(ageText,out int age)) { Console. ...
- 如何在HHDI中进行数据质量探查并获取数据剖析报告
通过执行多种数据剖析规则,对目标表(或一段SQL语句)进行数据质量探查,从而得到其数据质量情况.目前支持以下几种数据剖析类型,分别是:数字值分析.值匹配检查.字符值分析.日期值分析.布尔值分析.重复值 ...
- python django-admin startproject django-admin命令未找到
在使用pip install安装django后使用django-admin生成项目失败解决办法 1.配置环境变量-在系统环境变量path添加后运行 D:\Program Files (x86)\pyt ...
- 决策树&随机森林
参考链接: https://www.bilibili.com/video/av26086646/?p=8 <统计学习方法> 一.决策树算法: 1.训练阶段(决策树学习),也就是说:怎么样构 ...
- ruby中的return方法及class实例方法的initialize方法
return是函数的返回值 class Mtring def initialize(str) @name = str end def aa ary = @name.split(/:/) return ...
- HDU 5212 莫比乌斯反演
Code Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submis ...
- HttpClient的Content-Type设置
HttpClient的Content-Type设置 最近在对接公司内容的一个云服务的时候,遇到一个问题,就是如果使用HttpClient如何设置post时候的Content-Type? public ...