HDU 3338 Kakuro Extension(网络流)

因为某个地方m写成n玄学调bug一个小时...

写完内心是崩溃的...

#include<iostream>
#include<cstdio>
#include<cstring>
#include <list>
#include <algorithm>
#include <vector>
#include <queue>
#define N 40010
#define inf 1000000000
using namespace std;
class Graphic{
public:
    friend void show();
    void init(int a,int b){
        src = a,dst = b;
        if(dst >= N)for(;;);
        for(int i = 0 ; i <= dst ; ++i)G[i].clear();
        edge.clear();
        cnt = 0;
    }
    void add(int u,int v,int cap){
        _add(u,v,cap);
        _add(v,u,0);
    }
    int Dinic(){
        int ans = 0;
        while (bfs()) {
            memset(cur, 0, sizeof(cur));
            ans+=dfs(src,inf);
        }
        return ans;
    }
private:
    bool bfs(){
        while (!que.empty())que.pop();
        que.push(src);
        memset(dis,-1, sizeof(dis));
        dis[src] = 0;
        while(!que.empty()){
            int u = que.front();
            que.pop();
            for(int i = 0 ; i < G[u].size() ; ++i){
                int v = edge[G[u][i]].to;
                if(edge[G[u][i]].cap and dis[v]==-1){
                    dis[v] = dis[u]+1;
                    que.push(v);
                }
            }
        }
        return dis[dst]!=-1;
    }
    int dfs(int u,int a){
        if(u==dst or a == 0)return a;
        int ret = 0,f;
        for(int &i = cur[u] ; i < G[u].size() ; ++i){
            int v = edge[G[u][i]].to;
            int tmp = edge[G[u][i]].cap;
            if(dis[v] == dis[u] + 1 and tmp > 0 and (f = dfs(v,min(a,tmp)))>0){
                ret+=f;
                edge[G[u][i]].cap-=f;
                edge[G[u][i]^1].cap+=f;
                a-=f;
                if(a==0)break;
            }
        }
        return ret;
    }
private:
    int src,dst,dis[N],cur[N];
    int cnt;
    struct Edge{
        int to,cap;
        Edge(int a,int b):to(a),cap(b){}
    };
    queue<int>que;
    vector<int>G[N];
    vector<Edge>edge;
    void _add(int u,int v,int d){
        edge.emplace_back(Edge(v,d));
        G[u].push_back(cnt++);
    }
};
Graphic ss;
char str[105][105][10];
int down_right[105][105][2],n,m;
int get_dat(char *s){
    int ans = 0;
    while (*s and isdigit(*s)){
        ans = ans*10+*s-'0';
        ++s;
    }
    return ans;
}
int id[105][105],tol;
void deal(int i,int j){
    int cnt;
    down_right[i][j][0] = isdigit(str[i][j][0])?get_dat(str[i][j]):-1;
    down_right[i][j][1] = isdigit(str[i][j][4])?get_dat(str[i][j]+4):-1;
    if(isdigit(str[i][j][0])){
        cnt = 0;
        for(int k = i+1 ; k <= n and str[k][j][0]=='.' ; ++k){
            ++cnt;
        }
        down_right[i][j][0]-=cnt;
    }
    if(isdigit(str[i][j][4])){
        cnt = 0;
        for(int k = j+1 ; k <= m and str[i][k][0]=='.' ; ++k){
            ++cnt;
        }
        down_right[i][j][1]-=cnt;
    }
    if(isdigit(str[i][j][0])or isdigit(str[i][j][4])){
        id[i][j]=++tol;
        ++tol;
    }
}
int res[105][105];
void show(){
    for(int i = 1 ; i <= n ; ++i){
        for(int j = 1 ; j <= m/**因为这里笔者调了一个小时....**/ ; ++j){
            if(down_right[i][j][1]!=-1){
                for(int k = 1 ; str[i][j+k][0]=='.' and k < ss.G[id[i][j]+1].size() ; ++k){
                   /// cerr<<ss.edge[ss.G[id[i][j]+1][k]].to<<" "<<id[i][j]+1<<endl;
                    res[i][j+k] = 9 - ss.edge[ss.G[id[i][j]+1][k]].cap;
                }
            }
        }
    }
    for(int i = 1 ; i <= n ; ++i){
        for(int j = 1 ; j <= m ; ++j){
            if(str[i][j][0]=='.'){
                printf("%d",res[i][j]);
                ///if(res[i][j]==0x3f3f3f3f)for(;;);
            }
            else putchar('_');
            putchar(j==m?'\n':' ');
        }
        puts("");
    }
}
int main(){
    while(cin>>n>>m){
        for(int i = 1 ; i <= n ; ++i){
            for(int j = 1 ; j <= m ; ++j){
                scanf("%s",str[i][j]);
            }
        }
        int num = 0;
        tol = 0;
        for(int i = 1 ; i <= n ; ++i){
            for(int j = 1 ; j <= m ; ++j){
                deal(i,j);
            }
        }
        int src = 0,dst = tol+1;
        ss.init(src,dst);
        for(int i = 1 ; i <= n ; ++i){
            for(int j = 1 ; j <= m ; ++j){
                if(down_right[i][j][1]!=-1){
                ///    cerr<<i<<" "<<j<<":";
                    ss.add(src,id[i][j]+1,down_right[i][j][1]);
                    for(int k = j+1 ; k <= m and str[i][k][0]=='.' ; ++k){
                        int to = i-1;
                        for( ; to > 1 and str[to][k][0]=='.'; --to);
                        if(down_right[to][k][0]==-1)for(;;);
                        ss.add(id[i][j]+1,id[to][k],8);
                    }
                }
                if(down_right[i][j][0]!=-1){
                    ss.add(id[i][j],dst,down_right[i][j][0]);
                }
            }
        }
        ss.Dinic();
        show();
    }
}


/**
2 4
xxxxxxx 002/xxx xxxxxxx 002/xxx
xxx/002 ....... xxx/002 .......
 */

一个模板可以用一年..

从题目抽象出网络流的概念，从源点向横向格子建边，横向格子向对应竖向建边，竖向格子向汇点建边，跑网络流无脑找出流量即可。