标签：char fopen amp img 最大 nic size head second
题意：现在有一个n*m的矩形，现在每个白色的点都可以填 [1, 9] 中的一个数字。现在要求每行加起来的值等于左边的那个黑块的右值，每列加起来等于上边那个黑块的左值，求合法方案数。
题解：因为每个点至少是1，如果直接建边跑最大流的话会导致某些点的值为0，现在要保证每个点至少为1，我们可以直接把 s 流向每个白点流量为1， 然后黑点流入白点的流量减少边数的流量，最后跑最大流，输出答案。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb emplace_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define lch(x) tr[x].son[0] 12 #define rch(x) tr[x].son[1] 13 #define max3(a,b,c) max(a,max(b,c)) 14 #define min3(a,b,c) min(a,min(b,c)) 15 typedef pair<int,int> pll; 16 const int inf = 0x3f3f3f3f; 17 const LL INF = 0x3f3f3f3f3f3f3f3f; 18 const LL mod = (int)1e9+7; 19 const int N = 100*101 * 2; 20 const int M = N * 20; 21 int ans[150][150]; 22 int val[150][150][2]; 23 int head[N], deep[N], cur[N]; 24 int w[M], to[M], nx[M]; 25 int tot; 26 void add(int u, int v, int val){ 27 w[tot] = val; to[tot] = v; 28 nx[tot] = head[u]; head[u] = tot++; 29 30 w[tot] = 0; to[tot] = u; 31 nx[tot] = head[v]; head[v] = tot++; 32 } 33 int bfs(int s, int t){ 34 queue<int> q; 35 memset(deep, 0, sizeof(deep)); 36 q.push(s); 37 deep[s] = 1; 38 while(!q.empty()){ 39 int u = q.front(); 40 q.pop(); 41 for(int i = head[u]; ~i; i = nx[i]){ 42 if(w[i] > 0 && deep[to[i]] == 0){ 43 deep[to[i]] = deep[u] + 1; 44 q.push(to[i]); 45 } 46 } 47 } 48 return deep[t] > 0; 49 } 50 int Dfs(int u, int t, int flow){ 51 if(u == t) return flow; 52 for(int &i = cur[u]; ~i; i = nx[i]){ 53 if(deep[u]+1 == deep[to[i]] && w[i] > 0){ 54 int di = Dfs(to[i], t, min(w[i], flow)); 55 if(di > 0){ 56 w[i] -= di, w[i^1] += di; 57 return di; 58 } 59 } 60 } 61 return 0; 62 } 63 64 int Dinic(int s, int t){ 65 int ans = 0, tmp; 66 while(bfs(s, t)){ 67 for(int i = 0; i <= t; i++) cur[i] = head[i]; 68 while(tmp = Dfs(s, t, inf)) ans += tmp; 69 } 70 return ans; 71 } 72 void init(){ 73 memset(head, -1, sizeof(head)); 74 tot = 0; 75 } 76 char str[N]; 77 int n, m; 78 #define id(i,j) (i-1)*m+j 79 void GG(){ 80 for(int i = 1; i <= n; i++) 81 for(int j = 1; j <= m; j++){ 82 if(val[i][j][1] > 0){ 83 for(int z = head[id(i,j)]; ~z; z = nx[z]){ 84 if(z&1); 85 else { 86 int tx = to[z] / m + 1, ty = to[z] % m; 87 if(!ty) ty = m, tx -= 1; 88 ans[tx][ty] = 9 - w[z] + 1; 89 } 90 } 91 } 92 } 93 } 94 95 int main(){ 96 ///Fopen; 97 while(~scanf("%d%d", &n, &m)){ 98 init(); for(int i = 1; i <= n; i++) 99 for(int j = 1; j <= m; j++){ 100 ans[i][j] = 0; 101 scanf("%s", str+1); 102 if(str[1] == ‘.‘) val[i][j][0] = val[i][j][1] = -2; 103 else { 104 if(str[1] == ‘X‘) val[i][j][0] = -1; /// down 105 else val[i][j][0] = (str[1]-‘0‘)*100 + (str[2]-‘0‘)*10 + str[3] - ‘0‘; 106 if(str[5] == ‘X‘) val[i][j][1] = -1; 107 else val[i][j][1] = (str[5]-‘0‘)*100 + (str[6]-‘0‘)*10 + str[7] - ‘0‘; 108 } 109 } 110 int s = 0, t = n*m*2+1; 111 for(int i = 1; i <= n; i++){ 112 for(int j = 1; j <= m; j++){ 113 if(val[i][j][1] > 0){ 114 int tmp = val[i][j][1]; 115 int z = j+1; 116 while(z <= m && val[i][z][0] == -2){ 117 add(id(i,j), id(i,z), 9); 118 add(s, id(i,z), 1); 119 // cout << i << ‘ ‘ << j << " link " << i << ‘ ‘ << z << endl; 120 z++; 121 tmp--; 122 } 123 add(s, id(i,j), tmp); 124 j = z - 1; 125 } 126 } 127 } 128 for(int j = 1; j <= m; j++) 129 for(int i = 1; i <= n; i++){ 130 if(val[i][j][0] > 0){ 131 int z = i + 1; 132 add(id(i,j)+n*m, t, val[i][j][0]); 133 while(z <= n && val[z][j][0] == -2){ 134 add(id(z,j),id(i,j)+n*m,9); 135 z++; 136 } 137 i = z - 1; 138 } 139 } 140 //puts("Oh mather fuck!!!"); 141 int tt = Dinic(s,t); 142 //cout << tt << endl; 143 GG(); 144 for(int i = 1; i <= n; i++){ 145 for(int j = 1; j <= m; j++){ 146 if(ans[i][j]) printf("%d", ans[i][j]); 147 else printf("_"); 148 if(j!=m) printf(" "); 149 } 150 puts(""); 151 } 152 } 153 return 0; 154 } 155 /* 156 2 2 157 XXXXXXX 009/XXX 158 XXX/009 ....... 159 */
标签：char fopen amp img 最大 nic size head second
原文地址：https://www.cnblogs.com/MingSD/p/9738766.html
