首页 > 代码库 > hdu4888 Redraw Beautiful Drawings 最大流+判环
hdu4888 Redraw Beautiful Drawings 最大流+判环
hdu4888
Redraw Beautiful DrawingsTime Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2007 Accepted Submission(s): 447Problem 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 44 24 2 4 2 22 2 5 05 41 4 391 2 3 3 Sample Output Not UniqueImpossibleUnique1 2 3 3 Author Fudan University Source 2014 Multi-University Training Contest 3 Recommend We have carefully selected several similar problems for you: 4910 4909 4908 4907 4906 |
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。
（我的网络流是n个点，下标1~n的ISAP+GAP+CUR，怕不怕）
代码：
1 #include<cstdio> 2 #include<cmath> 3 #include<iostream> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 #include<map> 8 #include<set> 9 using namespace std; 10 #define ll __int64 11 #define usint unsigned int 12 #define RE freopen("1002.in","r",stdin) 13 #define WE freopen("1002my.out","w",stdout) 14 15 int ans; 16 int r[444],co[444]; 17 int k,nr,nc; 18 int sumr,sumc; 19 const int maxn=1111;//点数 20 const int maxm=444444;//边数 21 const int inf=0x7fffffff;//MAXINT 22 struct vnode { 23 int v,next; 24 int cap; 25 }; 26 int cnt,head[maxn]; 27 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 28 bool found; 29 int n,m,st,ed;//n个点m条边 30 int augc,flow;//augc为增广路容量，flow为最大流 31 vnode e[maxm]; 32 33 void add(int x,int y,int z) { 34 e[cnt].v=y; 35 e[cnt].cap=z; 36 e[cnt].next=head[x]; 37 head[x]=cnt; 38 cnt++; 39 // if(cnt>=maxm|| x>=maxn||y>=maxn) 40 // { 41 // //cout<<x<<‘,‘<<y<<‘,‘<<z<<‘,‘<<cnt<<‘,‘<<head[x]<<‘,‘<<head[y]<<endl; 42 // getchar(); 43 // } 44 e[cnt].v=x; 45 e[cnt].cap=0; 46 e[cnt].next=head[y]; 47 head[y]=cnt; 48 cnt++; 49 50 } 51 52 53 bool walked[maxn]; 54 bool dfs(int x,int preE) { 55 //cout<<x<<‘,‘; 56 for (int i=head[x]; i!=-1; i=e[i].next)//寻找容许边 57 { 58 if(i==(preE^1))continue; 59 if (e[i].cap>0) { //如果残留容量大于0，如果不是直接回头 60 if(walked[e[i].v])return true; 61 walked[e[i].v]=true; 62 if(dfs(e[i].v,i)) return true; 63 walked[e[i].v]=false; 64 } 65 } 66 //printf("(%d)out ",x); 67 //getchar(); 68 return false; 69 } 70 71 void aug(const int &m) { 72 int i,mini,minh=n-1; 73 int augco=augc; 74 if (m==ed) { //如果当前结点为汇点 75 found=true; 76 flow+=augc; //增加流量 77 return; 78 } 79 for (i=d[m]; i!=-1; i=e[i].next) { //寻找容许边 80 //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); 81 //getchar(); 82 if (e[i].cap && h[e[i].v]+1==h[m]) { //如果残留容量大于0，如果是容许边 83 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 84 d[m]=i; //把i定为当前弧 85 aug(e[i].v); //递归 86 if (h[st]>=n) return; //GAP 如果源点距离标号大于n 则停止算法 87 if (found) break; //如果找到汇点 则退出寻找 88 augc=augco;//没找到就还原当前的流 89 } 90 } 91 if (!found) { //重标号 92 for (i=head[m]; i!=-1; i=e[i].next) //找那个标号，这里不能用d[m]开始，不然会蛋疼 93 if (e[i].cap && h[e[i].v]<minh) { 94 minh=h[e[i].v]; 95 mini=i; 96 } 97 g[h[m]]--; //GAP 距离为 98 if (!g[h[m]]) h[st]=n; //GAP 99 h[m]=minh+1;100 d[m]=mini;101 g[h[m]]++; //GAP102 } else {103 //修改残量104 e[i].cap-=augc;105 e[i^1].cap+=augc;106 }107 }108 109 void farm() {110 int i,j,x,y,z;111 memset(head,-1,sizeof(head));112 cnt=0;113 n=nc+nr+2;114 st=n-1;115 ed=n;116 for(i=nc; i>=1; i--)117 add(st,i,co[i-1]);118 for(i=nr; i>=1; i--)119 add(nc+i,ed,r[i-1]);120 for(i=nc; i>=1; i--)121 for(j=nr; j>=1; j--)122 add(i,nc+j,k);123 //printf("cnt=%d\n",cnt);124 memset(h,0,sizeof(h));125 memset(g,0,sizeof(g));126 g[0]=n;127 flow=0;128 //printf("st=%d,head[st]=%d\n",st,head[st]);129 for(i=1; i<=n; i++)130 d[i]=head[i];//当前弧初始化131 while(h[st]<n) {132 augc=inf;//初始化增广路容量为正无穷大133 found=false;134 aug(st);//从源点开始找135 }136 if(flow!=sumr) {ans=0;return;}137 //printf("%d\n",flow);138 memset(walked,false,sizeof(walked));139 for(i=1;i<=nr;i++) ///Why is nr?140 {141 //printf("start dfs(%d)\n",i);142 if(dfs(i,-1)) {ans=2;return;}143 }144 ans=1;145 //printf("%d\n",flow);146 }147 148 int main() {149 //RE;150 //WE;151 int i,j;152 while(scanf("%d%d%d",&nr,&nc,&k)!=EOF) {153 sumr=0;sumc=0;154 for(i=0; i<nr; i++)155 {156 scanf("%d",&r[i]);157 sumr+=r[i];158 }159 for(i=0; i<nc; i++)160 {161 scanf("%d",&co[i]);162 sumc+=co[i];163 }164 ans=0;165 if(sumr==sumc)farm();166 if(ans==0) printf("Impossible\n");167 else if(ans!=1) {168 printf("Not Unique\n");169 // for(i=1; i<=nr; i++) {170 // for(j=head[nc+i]; j!=-1; j=e[j].next)171 // {172 // if(e[j].v==ed)continue;173 // printf("%d ",e[j].cap);174 // }175 // puts("");176 // }177 } else {178 printf("Unique\n");179 for(i=1; i<=nr; i++) {180 for(j=head[nc+i]; j!=-1; j=e[j].next) {181 if(e[j].v==ed)continue;182 printf("%d",e[j].cap);183 int thenext=e[j].next;184 while(thenext!=-1&&e[thenext].v==ed)thenext=e[thenext].next;185 if(thenext!=-1)putchar(‘ ‘);186 }187 puts("");188 }189 }190 }191 //cout<<"end";192 return 0;193 }
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