每个格子只能走有限次 求有多少人可以走出方格 每个人一次最多移动d格

#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
const int maxn = 1200;
const int INF = 0x3f3f3f3f;
struct{
int to,next,flow;
}e[maxn * maxn];
int S,T;
int head[maxn],cnt,d[maxn];
void add_edge(int from,int to,int flow){
e[cnt].to = to;
e[cnt].flow = flow;
e[cnt].next = head[from];
head[from] = cnt++;

e[cnt].to = from;
e[cnt].flow = 0;
e[cnt].next = head[to];
head[to] = cnt++;
}
int bfs(){
memset(d,0,sizeof(d));
queue<int>q;
q.push(S);
d[S] = 1;
while(!q.empty()){
int u = q.front();
q.pop();
for( int i = head[u]; i != -1; i = e[i].next){
int v = e[i].to;
if((!d[v]) && e[i].flow){
d[v] = d[u] + 1;
if(v == T) return 1;
q.push(v);
}
}
}
return 0;
}
int dfs(int u, int flow){
int cost = 0;
if(u == T) return flow;
for( int i = head[u]; i != -1; i = e[i].next){
int v = e[i].to;
if(d[v] == d[u] + 1 && e[i].flow){
int t = dfs(v,min(flow - cost, e[i].flow));
if(t){
e[i].flow -= t;
e[i^1].flow += t;
cost += t;
if(flow == cost) break;
}
}
}
return cost;
}
int Dinic(){
int res = 0;
while(bfs()){
res += dfs(S,INF);
}
return res;
}
char s1[1200][1200],s2[1200][1200];
int main(){
int T1,t1 = 1,n,d1;
scanf("%d",&T1);
while(T1--){
cnt = 0;
memset(head,-1,sizeof(head));
scanf("%d%d",&n,&d1);
for( int i = 1; i <= n; i++) scanf("%s",s1[i]+1);
for( int i = 1; i <= n; i++) scanf("%s",s2[i]+1);
int m = strlen(s1[1] + 1);
S = 2*n*m+1;
T = 2 * n * m + 2;
//	printf("1------------\n");
for(int i = 1; i <= n; i++){
for( int j = 1; j <= m; j++){
if(s1[i][j] - '0'){
add_edge((i - 1) * m + j,(i - 1) * m + j + n * m,s1[i][j] - '0');
//				printf("%d %d %d\n",(i - 1) * m + j,(i - 1) * m + j + n * m,s1[i][j] - '0');
}
}
}
int res = 0;
//		printf("2------------\n");
for( int i = 1; i <= n; i++){
for( int j = 1; j <= m; j++){
if(s2[i][j] == 'L'){
res++;
add_edge(S,(i - 1) * m + j,1);
//				printf("%d %d\n",S,(i-1)*m+j);
}
}
}
//		printf("3------------\n");
for( int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
if(i <= d1 || j <= d1 || j > m - d1 || i > n - d1){
add_edge( (i - 1) * m + j + n * m, T, INF);
//				printf("%d %d %d\n",(i - 1) * m + j + n * m,T,INF);
}
}
}
//	printf("4------------\n");
for (int i = 1;i <= n;++i) {
for (int j = 1;j <= m;++j) {
for (int l = 1;l <= n;++l) {
for (int r = 1;r <= m;++r) {
if (!(i==l && j==r) && abs(i-l) + abs(j-r) <= d1)
{
add_edge((i - 1) * m + j + n * m ,(l - 1) * m + r , INF);
//                     	printf("%d %d %d\n",(l - 1) * m + r,(i - 1) * m + j + n * m ,INF);}
}
}
}
}

}
printf("Case #%d: ",t1++);
int ans = Dinic();
if (res == ans) printf("no lizard was left behind.\n");
if (ans + 1 == res) printf("1 lizard was left behind.\n");
if (ans + 1 < res) printf("%d lizards were left behind.\n",res - ans);
}

}