Kakuro Extension【最大流】

HDU-3338

这道题真的处理起来好复杂啊，题意就是个简单的方格填数问题，但是每个白点至少放1，那么最后的可能解是怎样的呢？我们是不是要把x轴上的和y轴上的统一起来，然后就是每个点都被对应的x和y匹配起来，那么，之后，用每个点的x向y建立边，跑最大流，每个点的放的值就是反向边的权值了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(x, y) make_pair(x, y)
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e2 + 7, maxE = 1e5 + 7, st = 0;
pair<int, int> re_xy[20007];
int N, M, head[20007], cur[20007], cnt, ed, id[maxN][maxN], tot, ou[maxN][maxN];
char mp[maxN][maxN][10];
int down[maxN][maxN], righ[maxN][maxN];
struct Eddge
{
    int nex, to, flow;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), flow(c) {}
}edge[maxE];
inline void addEddge(int u, int v, int w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, int w) { addEddge(u, v, w); addEddge(v, u, 0); }
inline void solve(int x, int y)
{
    if(mp[x][y][1] == ‘X‘ && mp[x][y][7] == ‘X‘) { id[x][y] = -1; return; }
    if(mp[x][y][1] == ‘.‘) { return; }
    re_xy[++tot] = MP(x, y);
    id[x][y] = tot;
    down[x][y] = righ[x][y] = 0;
    if(mp[x][y][1] != ‘X‘) for(int i=1; i<=3; i++) down[x][y] = down[x][y] * 10 + mp[x][y][i] - ‘0‘;
    if(mp[x][y][5] != ‘X‘) for(int i=5; i<=7; i++) righ[x][y] = righ[x][y] * 10 + mp[x][y][i] - ‘0‘;
}
int deep[20007];
queue<int> Q;
inline bool bfs()
{
    for(int i=0; i<=ed; i++) deep[i] = 0;
    deep[st] = 1;
    while(!Q.empty()) Q.pop();
    Q.push(st);
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head[u], v, f; ~i; i=edge[i].nex)
        {
            v = edge[i].to; f = edge[i].flow;
            if(f && !deep[v])
            {
                deep[v] = deep[u] + 1;
                Q.push(v);
            }
        }
    }
    return deep[ed];
}
int  dfs(int u, int Dist)
{
    if(u == ed) return Dist;
    for(int &i=cur[u], v, f; ~i; i=edge[i].nex)
    {
        v = edge[i].to; f = edge[i].flow;
        if(f && deep[v] == deep[u] + 1)
        {
            int di = dfs(v, min(Dist, f));
            if(di)
            {
                edge[i].flow -= di;
                edge[i^1].flow += di;
                return di;
            }
        }
    }
    return 0;
}
inline int Dinic()
{
    int ans = 0, tmp;
    while(bfs())
    {
        for(int i=0; i<=ed; i++) cur[i] = head[i];
        while((tmp = dfs(st, INF))) ans += tmp;
    }
    return ans;
}
inline void init()
{
    cnt = tot = 0;
    memset(head, -1, sizeof(head));
    memset(id, 0, sizeof(id));
}
int main()
{
    while(scanf("%d%d", &N, &M) != EOF)
    {
        init();
        for(int i=1; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                scanf("%s", mp[i][j] + 1);
                solve(i, j);
            }
        }
        ed = (tot << 1) + 1;
        for(int i=1, le_id=0, up_id=0; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                if(mp[i][j][1] == ‘.‘)
                {
                    for(int dx = i - 1; dx > 0; dx--)
                    {
                        if(id[dx][j])
                        {
                            up_id = dx;
                            down[dx][j]--;
                            break;
                        }
                    }
                    for(int dy = j - 1; dy > 0; dy--)
                    {
                        if(id[i][dy])
                        {
                            le_id = dy;
                            righ[i][dy]--;
                            break;
                        }
                    }
                    _add(id[i][le_id], id[up_id][j] + tot, 8);
                }
            }
        }
        for(int i=1; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                if(id[i][j] > 0)
                {
                    if(mp[i][j][1] != ‘X‘) _add(id[i][j] + tot, ed, down[i][j]);
                    if(mp[i][j][5] != ‘X‘) _add(st, id[i][j], righ[i][j]);
                }
            }
        }
        Dinic();
        for(int x = 1; x <= tot; x++)
        {
            for(int i=head[x], y, f; ~i; i=edge[i].nex)
            {
                y = edge[i].to; f = edge[i^1].flow;
                if(y == st) continue;
                ou[re_xy[x].first][re_xy[y - tot].second] = f;
            }
        }
        for(int i=1; i<=N; i++)
        {
            for(int j=1; j<=M; j++)
            {
                if(mp[i][j][1] != ‘.‘) printf("_");
                else printf("%d", ou[i][j] + 1);
                printf("%c", j == M ? ‘\n‘ : ‘ ‘);
            }
        }
    }
    return 0;
}
/*
3 3
XXXXXXX 009/XXX XXXXXXX
XXX/009 ....... XXXXXXX
XXXXXXX XXXXXXX XXXXXXX
*/

原文地址：https://www.cnblogs.com/WuliWuliiii/p/11266490.html