多线程DP
Matrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2650 Accepted Submission(s): 1411
Problem Description
Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.
Every time yifenfei should to do is that choose a detour which frome the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix yifenfei choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And yifenfei can not pass the same area of the Matrix except the start and end.
Every time yifenfei should to do is that choose a detour which frome the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix yifenfei choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And yifenfei can not pass the same area of the Matrix except the start and end.
Input
The input contains multiple test cases.
Each case first line given the integer n (2<n<30)
Than n lines,each line include n positive integers.(<100)
Each case first line given the integer n (2<n<30)
Than n lines,each line include n positive integers.(<100)
Output
For each test case output the maximal values yifenfei can get.
Sample Input
2
10 3
5 10
3
10 3 3
2 5 3
6 7 10
5
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
Sample Output
28
46
80
Author
yifenfei
Source
Recommend
yifenfei
//费用流模板题，因为题目中的表述就是一个网络流模型，拆点建图，没点只经过一次，起点终点容量为2，
//其他点容量为1，要求最大费用费用为负权值，套模板。起点和终点经历了两次，最后要减去。
/****************最小费用最大流模板,白书363页*******************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=30*30*2+5,inf=0x7fffffff;//本体拆点，数组多开两倍
struct Edge
{
int from,to,cap,flow,cost;
Edge(int u,int v,int c,int f,int cs):from(u),to(v),cap(c),flow(f),cost(cs){}
};
struct MCMF
{
int n,m,s,t;
vector<Edge>edges;
vector<int>g[maxn];
int inq[maxn],d[maxn],p[maxn],a[maxn];
void init(int n)
{
this->n=n;
for(int i=0;i<n;i++) g[i].clear();
edges.clear();
}
void AddEdge(int from,int to,int cap,int cost)
{
edges.push_back((Edge){from,to,cap,0,cost});
edges.push_back((Edge){to,from,0,0,-cost});
m=edges.size();
g[from].push_back(m-2);
g[to].push_back(m-1);
}
bool BellmanFord(int s,int t,int &flow,int &cost)
{
for(int i=0;i<n;i++) d[i]=inf;
memset(inq,0,sizeof(inq));
d[s]=0;inq[s]=1;p[s]=0;a[s]=inf;
queue<int>q;
q.push(s);
while(!q.empty()){
int u=q.front();q.pop();
inq[u]=0;
for(int i=0;i<(int)g[u].size();i++){
Edge &e=edges[g[u][i]];
if(e.cap>e.flow&&d[e.to]>d[u]+e.cost){
d[e.to]=d[u]+e.cost;
p[e.to]=g[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to]) {q.push(e.to);inq[e.to]=1;}
}
}
}
if(d[t]==inf) return false;
flow+=a[t];
cost+=d[t]*a[t];
int u=t;
while(u!=s){
edges[p[u]].flow+=a[t];
edges[p[u]^1].flow-=a[t];
u=edges[p[u]].from;
}
return true;
}
int Mincost(int s,int t)
{
int flow=0,cost=0;
while(BellmanFord(s,t,flow,cost));
return cost;//返回最小费用，flow存最大流
}
}MC;
/**********************************************************************************/
int main()
{
int n,mp[35][35];
while(scanf("%d",&n)==1){
int s=0,t=n*n*2+1;
MC.init(n*n*2+2);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&mp[i][j]);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
int id=(i-1)*n+j;
if(id==1){
MC.AddEdge(id,id+n*n,2,-mp[i][j]);
MC.AddEdge(s,id,2,0);
}
else if(id==n*n){
MC.AddEdge(id,id+n*n,2,-mp[i][j]);
MC.AddEdge(id+n*n,t,2,0);
}
else MC.AddEdge(id,id+n*n,1,-mp[i][j]);
if(i<n){
int nid=id+n;
MC.AddEdge(id+n*n,nid,1,0);
}
if(j<n){
int nid=id+1;
MC.AddEdge(id+n*n,nid,1,0);
}
}
}
int ans=-(MC.Mincost(0,n*n*2+1)+mp[1][1]+mp[n][n]);
printf("%d\n",ans);
}
return 0;
}
Matrix Again
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 4255 Accepted Submission(s): 1233
Problem Description
Starvae very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.
Every time starvae should to do is that choose a detour which from the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix starvae choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And starvae can not pass the same area of the Matrix except the start and end..
Do you know why call this problem as “Matrix Again”? AS it is like the problem 2686 of HDU.
Every time starvae should to do is that choose a detour which from the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix starvae choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And starvae can not pass the same area of the Matrix except the start and end..
Do you know why call this problem as “Matrix Again”? AS it is like the problem 2686 of HDU.
Input
The input contains multiple test cases.
Each case first line given the integer n (2<=n<=600)
Then n lines, each line include n positive integers. (<100)
Each case first line given the integer n (2<=n<=600)
Then n lines, each line include n positive integers. (<100)
Output
For each test case output the maximal values starvae can get.
Sample Input
2 10 3 5 10 3 10 3 3 2 5 3 6 7 10 5 1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9
Sample Output
28 46 80
Author
Starvae
Source
//和HDU2686一样，只是数据变大了，上一个模板会超内存,这个板不会。
/***********************最小费用最大流模板2*************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=610*610*2+2;
const int maxm=4*maxn;//！边数要够
const int inf=0x7fffffff;
struct Edge
{
int to,next,cap,flow,cost;
}edges[maxm];
int head[maxn],tol,pre[maxn],dis[maxn];
bool vis[maxn];
int N;
void init(int n)
{
N=n;
tol=0;
memset(head,-1,sizeof(head));
}
void AddEdge(int u,int v,int cap,int cost)
{
edges[tol].to=v;
edges[tol].cap=cap;
edges[tol].cost=cost;
edges[tol].flow=0;
edges[tol].next=head[u];
head[u]=tol++;
edges[tol].to=u;
edges[tol].cap=0;
edges[tol].cost=-cost;
edges[tol].flow=0;
edges[tol].next=head[v];
head[v]=tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i=0;i<=N;i++){
dis[i]=inf;
vis[i]=0;
pre[i]=-1;
}
dis[s]=0;
vis[s]=1;
q.push(s);
while(!q.empty()){
int u=q.front();q.pop();
vis[u]=0;
for(int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
if(edges[i].cap>edges[i].flow&&dis[v]>dis[u]+edges[i].cost){
dis[v]=dis[u]+edges[i].cost;
pre[v]=i;
if(!vis[v]) {vis[v]=1;q.push(v);}
}
}
}
if(pre[t]==-1) return 0;
return 1;
}
int MinCostFlow(int s,int t)
{
int flow=0,cost=0;
while(spfa(s,t)){
int Min=inf;
for(int i=pre[t];i!=-1;i=pre[edges[i^1].to])
Min=min(Min,edges[i].cap-edges[i].flow);
for(int i=pre[t];i!=-1;i=pre[edges[i^1].to]){
edges[i].flow+=Min;
edges[i^1].flow-=Min;
cost+=edges[i].cost*Min;
}
flow+=Min;
}
return cost;//返回最小费用，flow存最大流
}
/*********************************************************************/
int main()
{
int n,mp[605][605];
while(scanf("%d",&n)==1){
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++) scanf("%d",&mp[i][j]);
int s=0,t=n*n*2+1;
init(n*n*2+2);
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
int id=(i-1)*n+j;
if(id==1){
AddEdge(id,id+n*n,2,-mp[i][j]);
AddEdge(s,id,2,0);
}
else if(id==n*n){
AddEdge(id,id+n*n,2,-mp[i][j]);
AddEdge(id+n*n,t,2,0);
}
else AddEdge(id,id+n*n,1,-mp[i][j]);
if(i<n) AddEdge(id+n*n,id+n,1,0);
if(j<n) AddEdge(id+n*n,id+1,1,0);
}
}
int ans=-(MinCostFlow(s,t)+mp[1][1]+mp[n][n]);
printf("%d\n",ans);
}
return 0;
}
给出代码供以后参考学习
永远渴望，大智若愚（stay hungry, stay foolish）
