扫二维码与项目经理沟通
我们在微信上24小时期待你的声音
解答本文疑问/技术咨询/运营咨询/技术建议/互联网交流
/*
最大流模板
isap复杂度(v^2*e)
*/
/*
*State: HDU4280 3609MS 8860K 3571 B C++
*题目大意:
* 有N个岛屿 M条无向路 每个路有一最大允许的客流量,求从最西的那个岛屿最多能运
* 用多少乘客到最东的那个岛屿。
*解题思路:
* 很单纯的网络流,重点是卡时间 模板的高效性很重要啊该模板详解
* 参见这里 模板题就不注释了
*解题感想:
* 10000ms的时间里,大多数最大流算法都挂掉了,就这个坚持下来了,用了3000ms左右。
*/
成都创新互联公司-专业网站定制、快速模板网站建设、高性价比玛沁网站开发、企业建站全套包干低至880元,成熟完善的模板库,直接使用。一站式玛沁网站制作公司更省心,省钱,快速模板网站建设找我们,业务覆盖玛沁地区。费用合理售后完善,十多年实体公司更值得信赖。
#include
#include
#include
#define VM 100010
#define EM 400010
using namespace std;
const int inf = 0x3f3f3f3f;
struct E
{
int to, frm, nxt, cap;
}edge[EM];
int head[VM], e;//建图时初始化
int dep[VM], gap[VM];//ISAP函数内初始化
void init()
{
e = 0;
memset(head, -1, sizeof(head));
}
void addEdge(int cu, int cv, int cw)
{
edge[e].frm = cu;
edge[e].to = cv;
edge[e].cap = cw;
edge[e].nxt = head[cu];
head[cu] = e++;
edge[e].frm = cv;
edge[e].to = cu;
edge[e].cap = 0;
edge[e].nxt = head[cv];
head[cv] = e++;
}
int que[VM];
void BFS(int des)
{
memset(dep, -1, sizeof(dep));
memset(gap, 0, sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[des] = 0;
que[rear++] = des;
int u, v;
while (front != rear)
{
u = que[front++];
front = front%VM;
for (int i=head[u]; i!=-1; i=edge[i].nxt)
{
v = edge[i].to;
if (edge[i].cap != 0 || dep[v] != -1)
continue;
que[rear++] = v;
rear = rear % VM;
++gap[dep[v] = dep[u] + 1];
}
}
}
int cur[VM], stack[VM];
//sap模板
int ISAP(int src, int des, int n)//源点、汇点、图中点的总数
{
int res = 0;
BFS(des);
int top = 0;
memcpy(cur, head, sizeof(head));
int u = src, i;
while (dep[src] < n)
{
if (u == des)
{
int temp = inf, inser = n;
for (i=0; i!=top; ++i)
if (temp > edge[stack[i]].cap)
{
temp = edge[stack[i]].cap;
inser = i;
}
for (i=0; i!=top; ++i)
{
edge[stack[i]].cap -= temp;
edge[stack[i]^1].cap += temp;
}
res += temp;
top = inser;
u = edge[stack[top]].frm;
}
if (u != des && gap[dep[u] -1] == 0)
break;
for (i = cur[u]; i != -1; i = edge[i].nxt)
if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1)
break;
if (i != -1)
{
cur[u] = i;
stack[top++] = i;
u = edge[i].to;
}
else
{
int min = n;
for (i = head[u]; i != -1; i = edge[i].nxt)
{
if (edge[i].cap == 0)
continue;
if (min > dep[edge[i].to])
{
min = dep[edge[i].to];
cur[u] = i;
}
}
--gap[dep[u]];
++gap[dep[u] = min + 1];
if (u != src)
u = edge[stack[--top]].frm;
}
}
return res;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
int n, m, T;
scanf("%d", &T);
while (T--)
{
scanf("%d %d", &n, &m);
int x, y, src, des;
int Min = inf, Max = -inf;
for (int i = 1; i <= n; ++i) //找出起点src 终点des
{
scanf("%d %d", &x, &y);
if (x <= Min)
{
src = i;
Min = x;
}
if (x >= Max)
{
des = i;
Max = x;
}
}
init();
int u, v, c;
for (int i = 0; i != m; ++i)
{
scanf("%d %d %d", &u, &v, &c);
addEdge(u, v, c);
addEdge(v, u, c);
}
int ans = ISAP(src, des, n);
printf("%d\n", ans);
}
return 0;
}
//Isap邻接矩阵版模板
//HDU3549 Isap邻接矩阵版
//46ms
#include
#include
#include
#define M 16
using namespace std;
const int inf = 0x3f3f3f3f;
int maze[M][M];
int gap[M],dis[M],pre[M],cur[M];
void init()
{
memset(maze, 0, sizeof(maze));
}
//起点是0, 终点是n-1,nodenum为点的个数
int ISAP(int s, int t, int nodenum) {
memset(cur, 0, sizeof(cur));
memset(dis, 0, sizeof(dis));
memset(gap, 0, sizeof(gap));
int u = pre[s] = s,maxflow = 0,aug = inf;
gap[0] = nodenum;
while(dis[s] < nodenum) {
loop:
for(int v = cur[u]; v < nodenum; v++) if(maze[u][v] && dis[u] == dis[v] + 1) {
aug = min(aug, maze[u][v]);
pre[v] = u;
u = cur[u] = v;
if(v == t) {
maxflow += aug;
for(u = pre[u];v != s;v = u,u = pre[u]) {
maze[u][v] -= aug;
maze[v][u] += aug;
}
aug = inf;
}
goto loop;
}
int mindis= nodenum-1;
for(int v = 0; v < nodenum; v++) if(maze[u][v] && mindis> dis[v]) {
cur[u] = v;
mindis= dis[v];
}
if((--gap[dis[u]])== 0) break;
gap[dis[u] = mindis+1] ++;
u = pre[u];
}
return maxflow;
}
int main(void)
{
#ifndef ONLINE_JUDGE
freopen("HDU3549.txt", "r", stdin);
#endif
int n, m, cas, cas_c = 1;
scanf("%d", &cas);
while(cas--)
{
scanf("%d %d", &n, &m);
init();
int u, v, w;
for(int i = 0; i < m; i++)
{
scanf("%d %d %d", &u, &v, &w);
u--, v--;
maze[u][v] += w;
}
int sol = ISAP(0, n-1, n);//起点,终点
printf("Case %d: %d\n", cas_c++, sol);
}
return 0;
}
//State: HDU4280居然用了2700ms+,牛逼模版
/*
This Problem beat me completely.
Beat my template completely.
Beat my funtion comletely.
Waste my time comletely.
It's a naked maximum-stream problem.
The main point is your template.
It make me realize that my SAP template is a sh*t!
....To make my SAP faster, I have done three kinds of optimization.
1. Record the adjacency node which has minimum distance;
2. Initailize the distance by using BFS;
3. As I got a Stack-Overflow, I simulate the DFS by using stack in STL;
Then I got AC in 2s+.
Of course, the problem is also correspond to the Minimum-Cut in Dual-Graph.
Hence, Dijkstra with minimum heap is optimal algorithm.
*/
//Header
//#include
//#include
//#include
//#include
//#include
//#include
//#include
//邻接表——sha
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define VM 10001
#define EM 500100
#define pi 3.1415926535897932
#define abs(x) ((x)>0?(x):-(x))
#define sqr(x) ((x)*(x))
#define fill(m,v) memset(m,v,sizeof(m))
#define SS(x) scanf("%d",&x)
#define inf 0x7fffffff
#define hinf 0x3f3f3f3f
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define rall(v) v.rbegin(),v.rend()
#define sz(v) ((int)v.size())
#define y0 stupid_cmath
#define y1 very_stupid_cmath
#define ll __int64
#define ull unsigned __int64
#define pb push_back
#define mp make_pair
#define rep(i,m) for(int i=0;i<(int)(m);i++)
#define rep2(i,n,m) for(int i=n;i<(int)(m);i++)
#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define FF(i,a) for(int i = 0; i < (a); i++ )
#define FFD(i,a) for(int i = (a)-1; i >= 0; i--)
#define CC(m,what) memset(m,what,sizeof(m))
#define SZ(a) ((int)a.size())
#define PP(n,m,a) puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
int dx[] = {-1, 0, 1, 0};//up Right down Left
int dy[] = {0, 1, 0, -1};
//#define For(it,c) for(__typeof(c.begin()) it=c.begin();it!=c.end();++it)
using namespace std;
#define M 40010
template
template
int gap[M],dis[M],pre[M],cur[M];
int NE,NV;
int head[M];
struct Node{
int c,pos,next;
}E[999999];
int sap(int s,int t) {
memset(dis,0,sizeof(int)*(NV+1));
memset(gap,0,sizeof(int)*(NV+1));
FF(i,NV) cur[i] = head[i];
int u = pre[s] = s,maxflow = 0,aug = -1;
gap[0] = NV;
while(dis[s] < NV) {
loop: for(int &i = cur[u]; i != -1; i = E[i].next) {
int v = E[i].pos;
if(E[i].c && dis[u] == dis[v] + 1) {
checkmin(aug,E[i].c);
pre[v] = u;
u = v;
if(v == t) {
maxflow += aug;
for(u = pre[u];v != s;v = u,u = pre[u]) {
E[cur[u]].c -= aug;
E[cur[u]^1].c += aug;
}
aug = -1;
}
goto loop;
}
}
int mindis = NV;
for(int i = head[u]; i != -1 ; i = E[i].next) {
int v = E[i].pos;
if(E[i].c && mindis > dis[v]) {
cur[u] = i;
mindis = dis[v];
}
}
if( (--gap[dis[u]]) == 0) break;
gap[ dis[u] = mindis+1 ] ++;
u = pre[u];
}
return maxflow;
}
void Insert(int u,int v,int c,int cc = 0) {
E[NE].c = c; E[NE].pos = v;
E[NE].next = head[u]; head[u] = NE++;
E[NE].c = cc; E[NE].pos = u;
E[NE].next = head[v]; head[v] = NE++;
}
int main() {
FF(i,NV) head[i] = -1;
NE = 0;
}
我们在微信上24小时期待你的声音
解答本文疑问/技术咨询/运营咨询/技术建议/互联网交流