Board Game
Accept: 54 Submit: 151
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of the board which is own by Fat brother is consisting of an integer 0. At each turn, he can choose two adjacent grids and add both the integer inside them by 1. But due to some unknown reason, the number of each grid can not be large than a given integer K. Also, Maze has already drown an N*M board with N*M integers inside each grid. What Fat brother would like to do is adding his board to be as same as Maze’s. Now we define the different value of two boards A and B as:
Now your task is to help Fat brother the minimal value of S he can get.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains three integers N, M and K which are mention above. Then N lines with M integers describe the board.
1 <= T <= 100, 1 <= N, M, K <= 9
0 <= the integers in the given board <= 9
Output
For each case, output the case number first, then output the minimal value of S Fat brother can get.
Sample Input
5
2 2 9
3 4
2 3
1 3 9
4 6 4
1 1 9
9
3 3 5
1 2 3
4 5 6
7 8 9
3 3 9
1 2 3
4 5 6
7 8 9
Sample Output
Case 1: 0
Case 2: 2
Case 3: 81
Case 4: 33
Case 5: 5
解题：转自http://blog.csdn.net/henryascend/article/details/38663589
建图什么的太不懂了
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<queue> 5 using namespace std; 6 struct Edge { 7 int from, to , cap, flow, cost , next; 8 }; 9 const int inf = 0x3f3f3f3f; 10 int n,m,K; 11 Edge edge[3010]; 12 int head[110]; 13 int dx[4]= {0,0,-1,1}; 14 int dy[4]= {-1,1,0,0}; 15 int map[11][11]; 16 int d[110],a[110],p[110]; 17 bool vis[110]; 18 int ans,cnt; 19 void add(int from, int to , int cap, int cost) { 20 edge[cnt].from = from; 21 edge[cnt].to = to; 22 edge[cnt].cap =cap; 23 edge[cnt].flow = 0; 24 edge[cnt].cost = cost; 25 edge[cnt].next = head[from]; 26 head[from] = cnt ++; 27 28 edge[cnt].from = to; 29 edge[cnt].to = from; 30 edge[cnt].cap = 0; 31 edge[cnt].flow = 0; 32 edge[cnt].cost = -cost; 33 edge[cnt].next = head[to]; 34 head[to] = cnt ++; 35 } 36 37 bool bound(int x,int y) { 38 return ( x>=1 && x<=n && y>=1 && y<=m ); 39 } 40 41 bool spfa(int S, int T, int &flow, int &cost) { 42 memset(d,63,sizeof(d)); 43 memset(vis, 0, sizeof(vis)); 44 queue <int> q; 45 d[S] = 0; 46 a[S] = inf; 47 vis[S] = 1; 48 p[S] = 0; 49 q.push(S); 50 while (!q.empty()) { 51 int u = q.front(); 52 q.pop(); 53 int i=head[u]; 54 while (i!=-1) { 55 int v= edge[i].to; 56 if (edge[i].cap>edge[i].flow && d[v]>d[u]+edge[i].cost) { 57 d[v] = d[u] +edge[i].cost; 58 p[v] = i; 59 a[v] = min( a[u], edge[i].cap - edge[i].flow); 60 if (!vis[v]) { 61 q.push(v); 62 vis[v] =1; 63 } 64 } 65 i = edge[i].next; 66 } 67 vis[u]=0; 68 } 69 if (d[T]>=0) return false; 70 flow += a[T]; 71 cost += d[T] * a[T]; 72 int u= T; 73 while ( u!=S ) { 74 edge[ p[u] ] .flow +=a[T]; 75 edge[ p[u]^1 ].flow -=a[T]; 76 u=edge[ p[u] ].from; 77 } 78 return true; 79 } 80 81 void Mincost (int S, int T) { 82 int flow = 0 , cost = 0; 83 while ( spfa(S, T, flow, cost) ); 84 ans += cost; 85 } 86 int main() { 87 int T,cas=0; 88 scanf("%d",&T); 89 while (T--) { 90 memset(head,-1,sizeof(head)); 91 cnt =ans = 0; 92 scanf("%d%d%d",&n,&m,&K); 93 int idx=0, x; 94 int st= 0, en = n*m+1; 95 for (int i=1; i<=n; i++) 96 for (int j=1; j<=m; j++) { 97 scanf("%d",&x); 98 ans += x * x; 99 map[i][j]= ++idx; 100 for (int k=1; k<=K; k++) 101 if ( i % 2 == j % 2) 102 add( st, map[i][j], 1 , 2*k -1 - 2*x ); 103 else 104 add( map[i][j],en , 1 , 2*k -1 - 2*x ); 105 } 106 for (int i=1; i<=n; i++) 107 for (int j=1; j<=m; j++) { 108 if ( i % 2 == j % 2 ) 109 for (int k=0; k<4; k++) { 110 int tx= i + dx[k]; 111 int ty= j + dy[k]; 112 if (!bound( tx,ty)) continue; 113 add(map[i][j], map[tx][ty],inf, 0); 114 } 115 } 116 Mincost(st, en); 117 printf("Case %d: %d ",++cas,ans); 118 } 119 return 0; 120 }