UVa 11082 & 最大流的行列模型
题意:
给出一个矩阵前i行的和与前j列的和,(i∈[1,r],j属于[1,c]),每个元素ai,j∈[1,20],请你还原出这个矩阵,保证有解.
SOL:
给网络流建模跪了,神一样的建图,如果我我会怎么做呢?...搜索?然而每个元素具有行,列双重相关性...暴力都打不出来吧...
然而我们顺着搜索的方向想,如果每个点的搜索值最终小于这行的和,那么我们应该做什么?增大它!是不是感觉有点增广的想法出来了------>然而我只是瞎BB...事后觉得可以这么想但考场上并不能这么想出来...
考虑它的建图,因为每个元素至少为1,即流量至少为1,那它就变成一个有上下界的网络流问题...不会怎么办?因为每个元素都大于等于1那么我们把他都减一!
我们对每一行的和对应一个点与超级源S相连,容量为这一行的和,对于每条列我们建一个点与汇点相连,容量为列的和,然后行与列的点两两相连,容量为19,也就是数的最大值,然后跑一遍最大流,行与列之间每条边的流量就是原矩阵中的元素.
如何证明它的正确性?
显然所有行与列的和是相等的,即源汇的出入边一定都满流,同时,我们可以把每一行与所有列的连边上的流量看做该行对此列所做的"贡献",因为最大流一定使源汇出入边满流,所以一定有一种方案.那么这个显然是正确的.
发现行与列的关系以及和的限制条件,也许是思路突破的关键吧.
警报!警报!警报!!
自己打的代码莫名TLE了...因为点不多所以老人家写的EK,然而我写的dinic居然TLE了?smg?...因为时间不多先贴上代码,以后有空回来看...模型是关健啊.
TLE代码:
/*==========================================================================
# Last modified: 2016-03-08 08:03
# Filename: uva11082.cpp
# Description:
==========================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <cmath>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1]
#define maxn 1000
#define maxm 1000000
#define pi 3.1415926535898
#define _e 2.718281828459
#define INF 1070000000
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template<class T> inline
void read(T& num) {
bool start=false,neg=false;
char c;
num=0;
while((c=getchar())!=EOF) {
if(c=='-') start=neg=true;
else if(c>='0' && c<='9') {
start=true;
num=num*10+c-'0';
} else if(start) break;
}
if(neg) num=-num;
}
/*==================split line==================*/
struct Edge{
int from,to,cap,v;
}e[maxm];
int r,c;
int id[maxn][maxn],a[maxn],b[maxn];
int first[maxn],next[maxm],cur[maxn],d[maxn];
bool vis[maxn];
int sume=1,n,S,T;
void addedge(int x,int y,int cap){
sume++; e[sume].from=x; e[sume].to=y; e[sume].cap=cap;
next[sume]=first[x]; first[x]=sume; id[x][y]=sume;
sume++; e[sume].from=y; e[sume].to=x; e[sume].cap=0;
next[sume]=first[y]; first[y]=sume; id[y][x]=sume;
}
void reset(){
sume=1;
a[0]=b[0]=0;
memset(e,0,sizeof(e));
memset(next,0,sizeof(next));
memset(first,0,sizeof(first));
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
//T=maxn-10,S=0;
}
bool bfs(){
queue<int>q;
FORP(i,S,T) d[i]=INF;
//memset(vis,false,sizeof(vis));
q.push(S); d[S]=0; //vis[S]=true;
while (!q.empty()){
int now=q.front(); q.pop();
for (int i=first[now];i;i=next[i])
if (d[e[i].to]==INF && e[i].cap){
d[e[i].to]=d[now]+1;
//vis[e[i].to]=true;
q.push(e[i].to);
}
}
return d[T]<INF;
}
int dfs(int now,int a){
if (now==T || !a) return a;
int f,flow=0;
for (int &i=cur[now];i;i=next[i])
if (d[now]+1==d[e[i].to] && (f=dfs(e[i].to,min(a,e[i].cap)))>0){
flow+=f; a-=f; e[i].cap-=f; e[i].v+=f; e[i^1].cap+=f;
if (!a) break;
}
return flow;
}
void dinic(){
int flow=0;
while (bfs()){
FORP(i,0,n) cur[i]=first[i];
flow+=dfs(S,INF);
}
}
int main(){
int cas; read(cas); //int cass=1;
FORP(cass,1,cas){
reset();
scanf("%d%d",&r,&c);
n=r+c+1; S=0; T=r+c+4;
FORP(i,1,r) scanf("%d",&a[i]);//read(a[i]);
FORP(i,1,c) scanf("%d",&b[i]);//read(b[i]);
FORP(i,1,r) addedge(0,i,a[i]-a[i-1]-c);
FORP(i,1,r)
FORP(j,r+1,r+c+1) addedge(i,j,19);
FORP(j,r+1,r+c+1) addedge(j,T,b[j-r]-b[j-r-1]-r);
printf("Matrix %d\n",cass); //cass++;
dinic();
FORP(i,1,r)
FORP(j,1,c)
printf("%d%c",e[id[i][j+r]].v+1,(j==c)?'\n':' ');
if (cas) cout << endl;
}
}
lrj代码:
// UVa11082 Matrix Decompressing
// Rujia Liu
// Slower version with EdmondsKarp
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 50 + 5;
const int INF = 1000000000;
struct Edge {
int from, to, cap, flow;
Edge(int u, int v, int c, int f):from(u),to(v),cap(c),flow(f) {}
};
struct EdmondsKarp {
int n, m;
vector<Edge> edges; // 边数的两倍
vector<int> G[maxn]; // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
int a[maxn]; // 当起点到i的可改进量
int p[maxn]; // 最短路树上p的入弧编号
void init(int n) {
for(int i = 0; i < n; i++) G[i].clear();
edges.clear();
}
void AddEdge(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
int Maxflow(int s, int t) {
int flow = 0;
for(;;) {
memset(a, 0, sizeof(a));
queue<int> Q;
Q.push(s);
a[s] = INF;
while(!Q.empty()) {
int x = Q.front(); Q.pop();
for(int i = 0; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if(!a[e.to] && e.cap > e.flow) {
p[e.to] = G[x][i];
a[e.to] = min(a[x], e.cap-e.flow);
Q.push(e.to);
}
}
if(a[t]) break;
}
if(!a[t]) break;
for(int u = t; u != s; u = edges[p[u]].from) {
edges[p[u]].flow += a[t];
edges[p[u]^1].flow -= a[t];
}
flow += a[t];
}
return flow;
}
};
EdmondsKarp g;
int no[maxn][maxn];
int main() {
int T, R, C, v, kase = 0;
scanf("%d", &T);
for(int kase = 1; kase <= T; kase++) {
scanf("%d%d", &R, &C);
g.init(R+C+2);
int last = 0;
for(int i = 1; i <= R; i++) {
scanf("%d", &v);
g.AddEdge(0, i, v - last - C); // row sum is v - last
last = v;
}
last = 0;
for(int i = 1; i <= C; i++) {
scanf("%d", &v);
g.AddEdge(R+i, R+C+1, v - last - R); // col sum is v - last
last = v;
}
for(int i = 1; i <= R; i++)
for(int j = 1; j <= C; j++) {
g.AddEdge(i, R+j, 19);
no[i][j] = g.edges.size() - 2; // no[i][j] is the index of arc for cell(i,j)
}
g.Maxflow(0, R+C+1);
printf("Matrix %d\n", kase);
for(int i = 1; i <= R; i++) {
for(int j = 1; j <= C; j++)
printf("%d ", g.edges[no[i][j]].flow + 1); // we subtracted 1 from every cell
printf("\n");
}
printf("\n");
}
return 0;
}
Sometimes it s the very people who no one imagines anything of. who do the things that no one can imagine.
