Redraw Beautiful DrawingsTime Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2007 Accepted Submission(s): 447 Problem Description
Alice and Bob are playing together. Alice is crazy about art and she has visited many museums around the world. She has a good memory and she can remember all drawings she has seen.
Today Alice designs a game using these drawings in her memory. First, she matches K+1 colors appears in the picture to K+1 different integers(from 0 to K). After that, she slices the drawing into grids and there are N rows and M columns. Each grid has an integer on it(from 0 to K) representing the color on the corresponding position in the original drawing. Alice wants to share the wonderful drawings with Bob and she tells Bob the size of the drawing, the number of different colors, and the sum of integers on each row and each column. Bob has to redraw the drawing with Alice's information. Unfortunately, somtimes, the information Alice offers is wrong because of Alice's poor math. And sometimes, Bob can work out multiple different drawings using the information Alice provides. Bob gets confused and he needs your help. You have to tell Bob if Alice's information is right and if her information is right you should also tell Bob whether he can get a unique drawing. Input
The input contains mutiple testcases.
For each testcase, the first line contains three integers N(1 ≤ N ≤ 400) , M(1 ≤ M ≤ 400) and K(1 ≤ K ≤ 40). N integers are given in the second line representing the sum of N rows. M integers are given in the third line representing the sum of M columns. The input is terminated by EOF. Output
For each testcase, if there is no solution for Bob, output "Impossible" in one line(without the quotation mark); if there is only one solution for Bob, output "Unique" in one line(without the quotation mark) and output an N * M matrix in the following N lines representing Bob's unique solution; if there are many ways for Bob to redraw the drawing, output "Not Unique" in one line(without the quotation mark).
Sample Input
2 2 4
4 2 4 2 4 2 2 2 2 5 0 5 4 1 4 3 9 1 2 3 3 Sample Output
Not Unique
Impossible Unique 1 2 3 3 Author
Fudan University
Source
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大意：n*m的矩阵，每个格子可以是0~k，给出各行的和和各列的和，求格子数字唯一方案，或判断无解或不唯一。
题解：
最大流，每行一个点，每列一个点，起点到每行的点连流量等于这行的和的边，每列的点连流量等于这列的和的边到终点，每行的点连到每列的点流量为K的点。所有行的和不等于所有列的和 或 最大流不等于所有行的和 则无解。有解的话再判有没有环，有环说明多解，可以dfs判环。
我这题卡了好久，判环的地方怒有问题，后来我看别人代码，只用从1~nr（行数）作为起始位置就能判完，也就是从行的节点开始，如果所有节点都开始一发的话会超时……我觉得再开点东西来记忆的话应该可以更快点，日后再说吧。
日后：
有了更好的删边判环方法，具体见一道几乎相同的题的题解：http://www.cnblogs.com/yuiffy/p/3929369.html
因为要删边判环，边被删了不好输出结果，所以先记录结果再删就好啦。速度快得飞起来。
（我的网络流是n个点，下标1~n的ISAP+GAP+CUR，怕不怕）
日前代码：
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
using namespace std;
#define ll __int64
#define usint unsigned int
#define RE freopen("1002.in","r",stdin)
#define WE freopen("1002my.out","w",stdout) int ans;
int r[],co[];
int k,nr,nc;
int sumr,sumc;
const int maxn=;//点数
const int maxm=;//边数
const int inf=0x7fffffff;//MAXINT
struct vnode {
int v,next;
int cap;
};
int cnt,head[maxn];
int h[maxn],g[maxn],d[maxn];//g[i]为标号为i的结点个数，h[i]为i结点的标号，d[]当前弧优化，记录当前弧
bool found;
int n,m,st,ed;//n个点m条边
int augc,flow;//augc为增广路容量，flow为最大流
vnode e[maxm]; void add(int x,int y,int z) {
e[cnt].v=y;
e[cnt].cap=z;
e[cnt].next=head[x];
head[x]=cnt;
cnt++;
// if(cnt>=maxm|| x>=maxn||y>=maxn)
// {
// //cout<<x<<','<<y<<','<<z<<','<<cnt<<','<<head[x]<<','<<head[y]<<endl;
// getchar();
// }
e[cnt].v=x;
e[cnt].cap=;
e[cnt].next=head[y];
head[y]=cnt;
cnt++; } bool walked[maxn];
bool dfs(int x,int preE) {
//cout<<x<<',';
for (int i=head[x]; i!=-; i=e[i].next)//寻找容许边
{
if(i==(preE^))continue;
if (e[i].cap>) { //如果残留容量大于0，如果不是直接回头
if(walked[e[i].v])return true;
walked[e[i].v]=true;
if(dfs(e[i].v,i)) return true;
walked[e[i].v]=false;
}
}
//printf("(%d)out ",x);
//getchar();
return false;
} void aug(const int &m) {
int i,mini,minh=n-;
int augco=augc;
if (m==ed) { //如果当前结点为汇点
found=true;
flow+=augc; //增加流量
return;
}
for (i=d[m]; i!=-; i=e[i].next) { //寻找容许边
//printf("m=%d,i=%d,e[i].v=%d,e[i].cap=%d,e[i].next=%d\n",m,i,e[i].v,e[i].cap,e[i].next);
//getchar();
if (e[i].cap && h[e[i].v]+==h[m]) { //如果残留容量大于0，如果是容许边
if (e[i].cap < augc) augc=e[i].cap;//如果容许边流量小于当前增广路流量 则更新增广路流量
d[m]=i; //把i定为当前弧
aug(e[i].v); //递归
if (h[st]>=n) return; //GAP 如果源点距离标号大于n 则停止算法
if (found) break; //如果找到汇点 则退出寻找
augc=augco;//没找到就还原当前的流
}
}
if (!found) { //重标号
for (i=head[m]; i!=-; i=e[i].next) //找那个标号，这里不能用d[m]开始，不然会蛋疼
if (e[i].cap && h[e[i].v]<minh) {
minh=h[e[i].v];
mini=i;
}
g[h[m]]--; //GAP 距离为
if (!g[h[m]]) h[st]=n; //GAP
h[m]=minh+;
d[m]=mini;
g[h[m]]++; //GAP
} else {
//修改残量
e[i].cap-=augc;
e[i^].cap+=augc;
}
} void farm() {
int i,j,x,y,z;
memset(head,-,sizeof(head));
cnt=;
n=nc+nr+;
st=n-;
ed=n;
for(i=nc; i>=; i--)
add(st,i,co[i-]);
for(i=nr; i>=; i--)
add(nc+i,ed,r[i-]);
for(i=nc; i>=; i--)
for(j=nr; j>=; j--)
add(i,nc+j,k);
//printf("cnt=%d\n",cnt);
memset(h,,sizeof(h));
memset(g,,sizeof(g));
g[]=n;
flow=;
//printf("st=%d,head[st]=%d\n",st,head[st]);
for(i=; i<=n; i++)
d[i]=head[i];//当前弧初始化
while(h[st]<n) {
augc=inf;//初始化增广路容量为正无穷大
found=false;
aug(st);//从源点开始找
}
if(flow!=sumr) {ans=;return;}
//printf("%d\n",flow);
memset(walked,false,sizeof(walked));
for(i=;i<=nr;i++) ///Why is nr?
{
//printf("start dfs(%d)\n",i);
if(dfs(i,-)) {ans=;return;}
}
ans=;
//printf("%d\n",flow);
} int main() {
//RE;
//WE;
int i,j;
while(scanf("%d%d%d",&nr,&nc,&k)!=EOF) {
sumr=;sumc=;
for(i=; i<nr; i++)
{
scanf("%d",&r[i]);
sumr+=r[i];
}
for(i=; i<nc; i++)
{
scanf("%d",&co[i]);
sumc+=co[i];
}
ans=;
if(sumr==sumc)farm();
if(ans==) printf("Impossible\n");
else if(ans!=) {
printf("Not Unique\n");
// for(i=1; i<=nr; i++) {
// for(j=head[nc+i]; j!=-1; j=e[j].next)
// {
// if(e[j].v==ed)continue;
// printf("%d ",e[j].cap);
// }
// puts("");
// }
} else {
printf("Unique\n");
for(i=; i<=nr; i++) {
for(j=head[nc+i]; j!=-; j=e[j].next) {
if(e[j].v==ed)continue;
printf("%d",e[j].cap);
int thenext=e[j].next;
while(thenext!=-&&e[thenext].v==ed)thenext=e[thenext].next;
if(thenext!=-)putchar(' ');
}
puts("");
}
}
}
//cout<<"end";
return ;
}
日后代码：
//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usll unsigned ll
#define mz(array) memset(array, 0, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) prllf("%d\n",x);
#define RE freopen("1002.in","r",stdin)
#define WE freopen("1002testout.txt","w",stdout)
#define mp make_pair
#define pb push_back int ans;
int r[],co[];
int k,nr,nc;
int sumr,sumc;
const int maxn=;//点数
const int maxm=;//边数
const int inf=0x7fffffff;//MAXINT
struct vnode {
int v,next;
int cap;
};
int cnt,head[maxn];
int h[maxn],g[maxn],d[maxn];//g[i]为标号为i的结点个数，h[i]为i结点的标号，d[]当前弧优化，记录当前弧
bool found;
int n,m,st,ed;//n个点m条边
int augc,flow;//augc为增广路容量，flow为最大流
vnode e[maxm]; int an[maxn][maxn]; void add(int x,int y,int z) {
e[cnt].v=y;
e[cnt].cap=z;
e[cnt].next=head[x];
head[x]=cnt;
cnt++;
// if(cnt>=maxm|| x>=maxn||y>=maxn)
// {
// //cout<<x<<','<<y<<','<<z<<','<<cnt<<','<<head[x]<<','<<head[y]<<endl;
// getchar();
// }
e[cnt].v=x; e[cnt].cap=;
e[cnt].next=head[y];
head[y]=cnt;
cnt++; } bool walked[maxn];
bool dfs(const int &x,const int &prex) {///深搜判环
int biu=-;
walked[x]=true;
for (int i=head[x]; i!=-; i=e[i].next) {
if(e[i].v==prex) {
biu=i;
continue;
}
if (e[i].cap>) {
if(walked[e[i].v]) return true;
if(dfs(e[i].v,x)) return true;
}
if(biu==-) head[x]=e[i].next;///删边，因为走了这条边却没发现环
else e[biu].next=e[i].next;
biu=i;
}
walked[x]=false;
return false;
} void aug(const int &m) {
int i,mini,minh=n-;
int augco=augc;
if (m==ed) { //如果当前结点为汇点
found=true;
flow+=augc; //增加流量
return;
}
for (i=d[m]; i!=-; i=e[i].next) { //寻找容许边
//printf("m=%d,i=%d,e[i].v=%d,e[i].cap=%d,e[i].next=%d\n",m,i,e[i].v,e[i].cap,e[i].next);
//getchar();
if (e[i].cap && h[e[i].v]+==h[m]) { //如果残留容量大于0，如果是容许边
if (e[i].cap < augc) augc=e[i].cap;//如果容许边流量小于当前增广路流量 则更新增广路流量
d[m]=i; //把i定为当前弧
aug(e[i].v); //递归
if (h[st]>=n) return; //GAP 如果源点距离标号大于n 则停止算法
if (found) break; //如果找到汇点 则退出寻找
augc=augco;//没找到就还原当前的流
}
}
if (!found) { //重标号
for (i=head[m]; i!=-; i=e[i].next) //找那个标号，这里不能用d[m]开始，不然会蛋疼
if (e[i].cap && h[e[i].v]<minh) {
minh=h[e[i].v];
mini=i;
}
g[h[m]]--; //GAP 距离为
if (!g[h[m]]) h[st]=n; //GAP
h[m]=minh+;
d[m]=mini;
g[h[m]]++; //GAP
} else {
//修改残量
e[i].cap-=augc;
e[i^].cap+=augc;
}
} void farm() {
int i,j,x,y,z;
memset(head,-,sizeof(head));
cnt=;
n=nc+nr+;
st=n-;
ed=n;
for(i=nc; i>=; i--)
add(st,i,co[i-]);
for(i=nr; i>=; i--)
add(nc+i,ed,r[i-]);
for(i=nc; i>=; i--)
for(j=nr; j>=; j--)
add(i,nc+j,k);
//printf("cnt=%d\n",cnt);
memset(h,,sizeof(h));
memset(g,,sizeof(g));
g[]=n;
flow=;
//printf("st=%d,head[st]=%d\n",st,head[st]);
for(i=; i<=n; i++)
d[i]=head[i];//当前弧初始化
while(h[st]<n) {
augc=inf;//初始化增广路容量为正无穷大
found=false;
aug(st);//从源点开始找
}
if(flow!=sumr) {
ans=;
return;
}
//printf("%d\n",flow);
for(i=; i<=nr; i++) {
int k=;
for(j=head[nc+i]; j!=-; j=e[j].next) {
if(e[j].v==ed)continue;
an[i][k++]=e[j].cap;
int thenext=e[j].next;
while(thenext!=-&&e[thenext].v==ed)thenext=e[thenext].next;
}
}
memset(walked,false,sizeof(walked));
for(i=nr; i>=; i--) { ///Why is nr?
//printf("start dfs(%d)\n",i);
if(dfs(i,-)) {
ans=;
return;
}
}
ans=;
//printf("%d\n",flow);
} int main() {
//RE;
//WE;
int i,j;
while(scanf("%d%d%d",&nr,&nc,&k)!=EOF) {
sumr=;
sumc=;
for(i=; i<nr; i++) {
scanf("%d",&r[i]);
sumr+=r[i];
}
for(i=; i<nc; i++) {
scanf("%d",&co[i]);
sumc+=co[i];
}
ans=;
if(sumr==sumc)farm();
if(ans==) printf("Impossible\n");
else if(ans!=) {
printf("Not Unique\n");
// for(i=1; i<=nr; i++) {
// for(j=head[nc+i]; j!=-1; j=e[j].next)
// {
// if(e[j].v==ed)continue;
// printf("%d ",e[j].cap);
// }
// puts("");
// }
} else {
printf("Unique\n");
for(i=; i<=nr; i++) {
if(nc>=)printf("%d",an[i][]);
for(j=; j<=nc; j++) {
printf(" %d",an[i][j]);
}
puts("");
}
}
}
//cout<<"end";
return ;
}