HDU3338 Kakuro Extension【网络流】

题意：往白格子中填数，\左下角的数代表向下连续的白格子数的和，右上角的数代表右边连续数的和

思路：最大流，源点 → 行和 → 白格子→ 列和 → 汇点。把有行和的格子看成一个点，连到右边的点，填1-9的数，我们都减一，容量为8，同样源点连到行和的点的容量也要减。列和的也一样

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<set>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;typedef pair<int,int> pii;const int maxn = 20005;struct edge {int to,cf,rev;};vector<edge> G[maxn];int lev[maxn],iter[maxn];struct data{int a,b,flag,num;}node[105][105];void init(int x,int y){for(int i = 0; i <= 2*x*y+3; i++)G[i].clear();for(int i = 0; i < x; i++){for(int j = 0; j < y; j++){node[i][j].a = 0;node[i][j].b = 0;node[i][j].flag = 0;node[i][j].num = 0;}}}void add(int from, int to, int cap){G[from].push_back((edge){to,cap,G[to].size()});G[to].push_back((edge){from,0,G[from].size()-1});}void bfs(int s){memset(lev,-1,sizeof lev);queue<int> q;lev[s] = 0;q.push(s);while(!q.empty()){int v = q.front();q.pop();for(int i = 0; i < G[v].size(); i++){edge &e = G[v][i];if(e.cf > 0 && lev[e.to] < 0){lev[e.to] = lev[v] + 1;q.push(e.to);}}}}int dfs(int v,int t, int f){if(v == t) return f;for(int &i = iter[v]; i < G[v].size(); i++){edge &e = G[v][i];if(e.cf > 0 && lev[v] < lev[e.to]){int d = dfs(e.to, t, min(f,e.cf));if(d > 0){e.cf -= d;G[e.to][e.rev].cf += d;return d;}}}return 0;}int maxflow(int s,int t){int flow = 0;while(1){bfs(s);if(lev[t] < 0) return flow;memset(iter,0,sizeof iter);int f;while((f = dfs(s,t,0x3f3f3f3f)) > 0)flow += f;}}int main(void){int n,m;char s[10];while(scanf("%d%d",&n,&m)!=EOF){init(n,m);for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){scanf("%s",s);if(s[0] == '.')node[i][j].flag = 1;else if(s[3] == '\\'){int sum = 0;if(s[0] != 'X'){for(int k = 0; k < 3; k++)sum = sum * 10 + s[k] - '0';node[i][j].a = sum;}if(s[4] != 'X'){sum = 0;for(int k = 4; k < 7; k++)sum = sum * 10 + s[k] - '0';node[i][j].b = sum;}}}}int st = 2*n*m, en = st+1;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(node[i][j].a){int x = i+1,num = 0;while(x < n && node[x][j].flag){add(x*m+j,i*m+j+n*m,8);x++;num++;}add(i*m+j+n*m,en,node[i][j].a-num);}if(node[i][j].b){int y = j+1,num = 0;while(y < m && node[i][y].flag){add(i*m+j,i*m+y,8);y++;num++;}add(st,i*m+j,node[i][j].b-num);}}}maxflow(st,en);for(int i = 0; i < G[st].size(); i++){int x = G[st][i].to;for(int j = 0; j < G[x].size(); j++){int y = G[x][j].to;if(y != st)node[y/m][y%m].num = 8 - G[x][j].cf + 1;}}for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(node[i][j].num)printf("%d",node[i][j].num);elseputchar('_');if(j == m-1)putchar('\n');elseputchar(' ');}}}return 0;}