[kuangbin带你飞]专题十一网络流\L HDU 3338 Kakuro Extension_kakuro extension hdu

本文链接：https://blog.youkuaiyun.com/Adolphrocs/article/details/86699658

题目描述	If you solved problem like this, forget it.Because you need to use a completely different algorithm to solve the following one. Kakuro puzzle is played on a grid of "black" and "white" cells. Apart from the top row and leftmost column which are entirely black, the grid has some amount of white cells which form "runs" and some amount of black cells. "Run" is a vertical or horizontal maximal one-lined block of adjacent white cells. Each row and column of the puzzle can contain more than one "run". Every white cell belongs to exactly two runs — one horizontal and one vertical run. Each horizontal "run" always has a number in the black half-cell to its immediate left, and each vertical "run" always has a number in the black half-cell immediately above it. These numbers are located in "black" cells and are called "clues".The rules of the puzzle are simple: 1.place a single digit from 1 to 9 in each "white" cell 2.for all runs, the sum of all digits in a "run" must match the clue associated with the "run" Given the grid, your task is to find a solution for the puzzle. 　　　　　　　　　　 Picture of the first sample input　　　　　　　　　　　　　　　　 Picture of the first sample output
Input	The first line of input contains two integers n and m (2 ≤ n,m ≤ 100) — the number of rows and columns correspondingly. Each of the next n lines contains descriptions of m cells. Each cell description is one of the following 7-character strings: .......— "white" cell; XXXXXXX— "black" cell with no clues; AAA\BBB— "black" cell with one or two clues. AAA is either a 3-digit clue for the corresponding vertical run, or XXX if there is no associated vertical run. BBB is either a 3-digit clue for the corresponding horizontal run, or XXX if there is no associated horizontal run. The first row and the first column of the grid will never have any white cells. The given grid will have at least one "white" cell.It is guaranteed that the given puzzle has at least one solution.
Output	Print n lines to the output with m cells in each line. For every "black" cell print '_' (underscore), for every "white" cell print the corresponding digit from the solution. Delimit cells with a single space, so that each row consists of 2m-1 characters.If there are many solutions, you may output any of them.
Sample Input	6 6 XXXXXXX XXXXXXX 028\XXX 017\XXX 028\XXX XXXXXXX XXXXXXX 022\022 ....... ....... ....... 010\XXX XXX\034 ....... ....... ....... ....... ....... XXX\014 ....... ....... 016\013 ....... ....... XXX\022 ....... ....... ....... ....... XXXXXXX XXXXXXX XXX\016 ....... ....... XXXXXXX XXXXXXX 5 8 XXXXXXX 001\XXX 020\XXX 027\XXX 021\XXX 028\XXX 014\XXX 024\XXX XXX\035 ....... ....... ....... ....... ....... ....... ....... XXXXXXX 007\034 ....... ....... ....... ....... ....... ....... XXX\043 ....... ....... ....... ....... ....... ....... ....... XXX\030 ....... ....... ....... ....... ....... ....... XXXXXXX
Sample Output	_ _ _ _ _ _ _ _ 5 8 9 _ _ 7 6 9 8 4 _ 6 8 _ 7 6 _ 9 2 7 4 _ _ _ 7 9 _ _ _ _ _ _ _ _ _ _ _ 1 9 9 1 1 8 6 _ _ 1 7 7 9 1 9 _ 1 3 9 9 9 3 9 _ 6 7 2 4 9 2 _

/*
网络流
	源 -> 列和 -> 单元 ->行和 ->汇 
*/ 
#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define Del(a,b) memset(a,b,sizeof(a))
const int N = 20500;
const int inf = 0x3f3f3f3f;
int n, m;
struct Node {
	int from, to, cap, flow;
};
vector<int> v[N];
vector<Node> e;
int vis[N];  //构建层次图
int cur[N];
void add_Node(int from, int to, int cap) {
	e.push_back((Node) {from, to, cap, 0});
	e.push_back((Node) {to, from, 0, 0});
	int tmp = e.size();
	v[from].push_back(tmp - 2);
	v[to].push_back(tmp - 1);
}
bool bfs(int s, int t) {
	Del(vis, -1);
	queue<int> q;
	q.push(s);
	vis[s] = 0;
	while(!q.empty()) {
		int x = q.front();
		q.pop();
		for(int i = 0; i<v[x].size(); i++) {
			Node tmp = e[v[x][i]];
			if(vis[tmp.to]<0 && tmp.cap>tmp.flow) { //第二个条件保证
				vis[tmp.to]=vis[x]+1;
				q.push(tmp.to);
			}
		}
	}
	if(vis[t]>0)
		return true;
	return false;
}
int dfs(int o,int f,int t) {
	if(o==t || f==0)  //优化
		return f;
	int a = 0,ans=0;
	for(int &i=cur[o]; i<v[o].size(); i++) { //注意前面 ’&‘,很重要的优化
		Node &tmp = e[v[o][i]];
		if(vis[tmp.to]==(vis[o]+1) && (a = dfs(tmp.to,min(f,tmp.cap-tmp.flow),t))>0) {
			tmp.flow+=a;
			e[v[o][i]^1].flow-=a; //存图方式
			ans+=a;
			f-=a;
			if(f==0)  //注意优化
				break;
		}
	}
	return ans;  //优化
}

int dinci(int s,int t) {
	int ans=0;
	while(bfs(s,t)) {
		Del(cur,0);
		int tm = dfs(s,inf,t);
		ans += tm;
	}
	return ans;
}
int mp[250][250];
struct Node1 {
	int x, y, z;
};
Node1 raw[N], col[N];
int solve(int raw_cnt, int id,int s) {
	int cnt = 0, pos = id + raw_cnt;
	return e[v[pos][1]].flow + 1;
}
int main() {
	//freopen("Input.txt","r",stdin);
	while(~scanf("%d%d",&n,&m)) {
		memset(mp,-1,sizeof(mp));
		memset(col,0,sizeof(col));
		memset(raw,0,sizeof(raw));
		int cnt=0,raw_cnt=0,col_cnt=0;
		for(int i=1; i<=n; i++) {
			for(int j=1; j<=m; j++) {
				string str;
				cin>>str;
				if(str[0] == '.') {
					mp[i][j] = ++cnt;
				} else {
					mp[i][j] = -1;
					if(str[0] != 'X') {
						int tmp = (str[0]-'0')*100 + (str[1]-'0')*10 + str[2]-'0';
						col[++col_cnt].x = i;
						col[col_cnt].y = j;
						col[col_cnt].z = tmp;
					}
					if(str[4] != 'X') {
						int tmp = (str[4]-'0')*100 + (str[5]-'0')*10 + str[6]-'0';
						raw[++raw_cnt].x = i;
						raw[raw_cnt].y = j;
						raw[raw_cnt].z = tmp;
					}
				}
			}
		}
		//printf("%d %d %d\n",raw_cnt,col_cnt,cnt);
		int start=0,t=col_cnt+cnt+raw_cnt+2;
		for(int i = 1; i <= raw_cnt; i++) {
			int x = raw[i].x;
			int y = raw[i].y;
			int cnt_len = 0;
			for(y = y+1; y <= m; y++) {
				if(mp[x][y] != -1) {
					cnt_len++;
					add_Node(i,raw_cnt+mp[x][y],8);
				} else   break;
			}
			add_Node(start,i,raw[i].z-cnt_len);
		}
		for(int i = 1; i <= col_cnt; i++) {
			int x = col[i].x;
			int y = col[i].y;
			int cnt_len = 0;
			for(x = x+1; x <= n; x++) {
				if(mp[x][y] != -1) {
					cnt_len++;
					add_Node(raw_cnt + mp[x][y],raw_cnt + cnt + i, 8);
				} else    break;
			}
			add_Node(raw_cnt + cnt + i, t, col[i].z-cnt_len);
		}
		int ans=dinci(start,t);
		for(int i=1; i<=n; i++) {
			for(int j=1; j<=m; j++) {
				if(mp[i][j]==-1)
					printf("_");
				else
					printf("%d",solve(raw_cnt,mp[i][j],cnt));
				printf("%c",j==m?'\n':' ');
			}
		}
		for(int i=0; i<=t; i++)
			v[i].clear();
		e.clear();
	}
	return 0;
}