标签：des style http color os io for ar
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4975
3 1 1 5 5 2 2 0 10 0 10 2 2 2 2 2 2
Case #1: So simple! Case #2: So naive! Case #3: So young!
题意：
给出每行每列的和，问是否存在这样的表格；每个小格放的数字只能是0--9。
代码如下：（套用别人HDU4888的模板）
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
#define ll __int64
#define eps 1e-8
const ll Mod=(1e9+7);
const int maxn = 510;
const int maxm = 50100;
int n,m,k;
int r[maxn],c[maxn];
int ma[maxn][maxn];
const int maxnode = 10000 + 5;
const int maxedge = 2000000 + 5;
const int oo = 1000000000;
int node, src, dest, nedge;
int head[maxnode], point[maxedge], next1[maxedge], flow[maxedge], capa[maxedge];//point[x]==y表示第x条边连接y，head，next为邻接表，flow[x]表示x边的动态值，capa[x]表示x边的初始值
int dist[maxnode], Q[maxnode], work[maxnode];//dist[i]表示i点的等级
void init(int _node, int _src, int _dest){//初始化，node表示点的个数，src表示起点，dest表示终点
node = _node;
src = _src;
dest = _dest;
for (int i = 0; i < node; i++) head[i] = -1;
nedge = 0;
}
void addedge(int u, int v, int c1, int c2){//增加一条u到v流量为c1，v到u流量为c2的两条边
point[nedge] = v, capa[nedge] = c1, flow[nedge] = 0, next1[nedge] = head[u], head[u] = (nedge++);
point[nedge] = u, capa[nedge] = c2, flow[nedge] = 0, next1[nedge] = head[v], head[v] = (nedge++);
}
bool dinic_bfs(){
memset(dist, 255, sizeof (dist));
dist[src] = 0;
int sizeQ = 0;
Q[sizeQ++] = src;
for (int cl = 0; cl < sizeQ; cl++)
for (int k = Q[cl], i = head[k]; i >= 0; i = next1[i])
if (flow[i] < capa[i] && dist[point[i]] < 0){
dist[point[i]] = dist[k] + 1;
Q[sizeQ++] = point[i];
}
return dist[dest] >= 0;
}
int dinic_dfs(int x, int exp){
if (x == dest) return exp;
for (int &i = work[x]; i >= 0; i = next1[i]){
int v = point[i], tmp;
if (flow[i] < capa[i] && dist[v] == dist[x] + 1 && (tmp = dinic_dfs(v, min(exp, capa[i] - flow[i]))) > 0){
flow[i] += tmp;
flow[i^1] -= tmp;
return tmp;
}
}
return 0;
}
int dinic_flow(){
int result = 0;
while (dinic_bfs()){
for (int i = 0; i < node; i++) work[i] = head[i];
while (1){
int delta = dinic_dfs(src, oo);
if (delta == 0) break;
result += delta;
}
}
return result;
}
//建图前,运行一遍init()；
//加边时，运行addedge(a,b,c,0),表示点a到b流量为c的边建成（注意点序号要从0开始）
//求解最大流运行dinic_flow(),返回值即为答案
bool judge(int sumrow){
int flow = 1,cost = 0;
for(int i = 1;i <= n;i++)
for(int j = n+1;j <= n+m;j ++)
addedge(i,j,k,0);
flow=dinic_flow();
if(flow != sumrow)
return false;
return true;
}
int main(){
int t,cas=1;;
scanf("%d",&t);
while(t--)
{
printf("Case #%d: ",cas++);
scanf("%d%d",&n,&m);
k=9;
init(n+m+2,0,n+m+1);
int flag = 0;
int sumrow = 0,colrow = 0;
for(int i = 1;i <= n;i++){
scanf("%d",&r[i]);
addedge(0,i,r[i],0);
sumrow += r[i];
if(r[i]<0 || r[i]>m*k)
flag = 1;
}
for(int j = 1;j <= m;j ++){
scanf("%d",&c[j]);
addedge(j+n,n+m+1,c[j],0);
colrow += c[j];
if(c[j]<0 || c[j]>n*k)
flag = 1;
}
if(sumrow != colrow){
printf("So naive!\n");
continue;
}
if(!judge(sumrow))
flag = 1;
if(flag == 1){
printf("So naive!\n");
continue;
}
memset(ma,-1,sizeof(ma));
int i,j;
for(i=1;i<=n;i++)
if(r[i]==0)
for(j=1;j<=m;j++)
ma[i][j]=0;
for(j=1;j<=m;j++)
if(c[j]==0)
for(i=1;i<=n;i++)
ma[i][j]=0;
int tt=2;
int sum,num,temp;
while(tt--)
{
for(i=1;i<=n;i++)
{
if(r[i]==0)
{
for(j=1;j<=m;j++)
if(ma[i][j]==-1)
ma[i][j]=0;
continue;
}
sum=0;
num=0;
for(j=1;j<=m;j++)
{
if(ma[i][j]==-1)
{
num++;
temp=j;
sum+=min(k,c[j]);
}
}
if(num==1)
{
ma[i][temp]=r[i];
r[i]-=ma[i][temp];
c[temp]-=ma[i][temp];
continue;
}
else if(sum==r[i])
{
for(j=1;j<=m;j++)
{
if(ma[i][j]==-1)
{
ma[i][j]=min(k,c[j]);
r[i]-=ma[i][j];
c[j]-=ma[i][j];
}
}
}
}
for(j=1;j<=m;j++)
{
if(c[j]==0)
{
for(i=1;i<=n;i++)
if(ma[i][j]==-1)
ma[i][j]=0;
continue;
}
sum=0;
num=0;
for(i=1;i<=n;i++)
{
if(ma[i][j]==-1)
{
num++;
temp=i;
sum+=min(k,r[i]);
}
}
if(num==1)
{
ma[temp][j]=c[j];
r[temp]-=ma[temp][j];
c[j]-=ma[temp][j];
continue;
}
else if(sum==c[j])
{
for(i=1;i<=n;i++)
{
if(ma[i][j]==-1)
{
ma[i][j]=min(k,r[i]);
r[i]-=ma[i][j];
c[j]-=ma[i][j];
}
}
}
}
}
flag=0;
for(i=1;i<=n;i++)
if(r[i]!=0)
{
flag=1;
break;
}
for(j=1;j<=m;j++)
if(c[j]!=0)
{
flag=1;
break;
}
if(flag==1)
printf("So young!\n");
else
{
printf("So simple!\n");
}
}
return 0;
}hdu 4975 A simple Gaussian elimination problem.（网络流，判断矩阵是否存在）,布布扣,bubuko.com
hdu 4975 A simple Gaussian elimination problem.（网络流，判断矩阵是否存在）
标签：des style http color os io for ar
原文地址：http://blog.csdn.net/u012860063/article/details/38736763
