Problem 2143 Board GameAccept: 128 Submit: 327
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
Sample Output
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <list> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d ", a) #define plld(a) printf("%lld ", a) #define pc(a) printf("%c ", a) #define ps(a) printf("%s ", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 1100, INF = 0x7fffffff; int n, m, s, t; int head[maxn], nex[maxn << 1], f[maxn], d[maxn], p[maxn], vis[maxn]; int cnt, value, flow; int dis[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}; struct node { int u, v, w, c; }Node[maxn << 1]; void add_(int u, int v, int w, int c) { Node[cnt].u = u; Node[cnt].v = v; Node[cnt].w = w; Node[cnt].c = c; nex[cnt] = head[u]; head[u] = cnt++; } void add(int u, int v, int w, int c) { add_(u, v, w, c); add_(v, u, -w, 0); } bool spfa() { for(int i= 0; i < maxn; i++) d[i] = INF; deque<int> Q; mem(vis, 0); mem(p, -1); Q.push_front(s); vis[s] = 1; p[s] = d[s] = 0; f[s] = INF; while(!Q.empty()) { int u = Q.front(); Q.pop_front(); vis[u] = 0; for(int i = head[u]; i != -1; i = nex[i]) { int v = Node[i].v; if(d[v] > d[u] + Node[i].w && Node[i].c > 0) { d[v] = d[u] + Node[i].w; p[v] = i; f[v] = min(Node[i].c, f[u]); if(!vis[v]) { if(Q.empty()) Q.push_back(v); else if(d[v] < d[Q.front()]) Q.push_front(v); else Q.push_back(v); vis[v] = 1; } } } } if(d[t] >= 0) return 0; flow += f[t]; value += d[t] * f[t]; for(int i = t; i != s; i = Node[p[i]].u) { Node[p[i]].c -= f[t]; Node[p[i] ^ 1].c += f[t]; } return 1; } void init() { mem(head, -1); cnt = 0; value = flow = 0; } void max_flow() { while(spfa()); pd(value); } int main() { int T, K, kase = 0; rd(T); while(T--) { init(); rd(n), rd(m), rd(K); s = 0, t = n * m + 1; for(int i = 1; i <= n; i++) { for(int j = 1; j <= m; j++) { int tmp; rd(tmp); value += tmp * tmp; for(int k = 1; k <= K; k++) if((i + j) & 1) add(s, (i - 1) * m + j, (k - tmp) * (k - tmp) - (k - 1 - tmp) * (k - 1 - tmp), 1); else add((i - 1) * m + j, t, (k - tmp) * (k - tmp) - (k - 1 - tmp) * (k - 1 - tmp), 1); if(( i + j) & 1) for(int k = 0; k < 4; k++) { int nx = i + dis[k][0]; int ny = j + dis[k][1]; if(nx < 1 || ny < 1 || nx > n || ny > m) continue; add((i - 1) * m + j, (nx - 1) * m + ny, 0, INF); } } } printf("Case %d: ", ++kase); max_flow(); } return 0; }