【Luogu】P3356火星探险问题（费用流）

　　题目链接

　　网络流一条边都不能多连？没道理呀？

　　不过单看这题的确是个sb题……

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<queue>
#define maxn 100
#define maxm 100000
#define lim n*m
#define F(i,j) ((i-1)*m+j)
#define T(i,j) (F(i,j)+lim)
using namespace std;
inline long long read(){
    long long num=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch==‘-‘)    f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*10+ch-‘0‘;
        ch=getchar();
    }
    return num*f;
}

struct Edge{
    int from,next,to,val,dis,flow;
}edge[maxm];
int head[maxn*maxn],num;
inline void addedge(int from,int to,int val,int dis){
    edge[++num]=(Edge){from,head[from],to,val,dis,0};
    head[from]=num;
}
inline void add(int from,int to,int val,int dis){
    addedge(from,to,val,dis);
    addedge(to,from,0,-dis);
}

inline int count(int i){    return i&1?i+1:i-1;    }

int dis[maxn*maxn];
int flow[maxn*maxn];
int pre[maxn*maxn];
bool vis[maxn*maxn];
int Start,End;
int n,m;

int spfa(){
    memset(dis,-127/3,sizeof(dis));    dis[Start]=0;    flow[Start]=0x7fffffff;
    queue<int>q;    q.push(Start);
    while(!q.empty()){
        int from=q.front();    q.pop();    vis[from]=0;
        for(int i=head[from];i;i=edge[i].next){
            int to=edge[i].to;
            if(edge[i].val<=edge[i].flow||dis[to]>=dis[from]+edge[i].dis)    continue;
            dis[to]=dis[from]+edge[i].dis;
            pre[to]=i;    flow[to]=min(flow[from],edge[i].val-edge[i].flow);
            if(vis[to])    continue;
            vis[to]=1;    q.push(to);
        }
    }
    int now=End;
    while(now!=Start){
        int ret=pre[now];
        edge[ret].flow+=flow[End];
        edge[count(ret)].flow-=flow[End];
        now=edge[ret].from;
    }
    return dis[End];
}

struct Node{
    int x,y;
};

Node calc(int ret){
    Node ans;
    ans.x=(ret-1)/m+1;
    ans.y=ret-(ans.x-1)*m;
    return ans;
}

Node q[maxn*maxn];int tot;

void dfs(int x,int y,int now){
    if(now==End)    return;
    for(int i=head[now];i;i=edge[i].next){
        int to=edge[i].to;
        if(edge[i].flow==0)    continue;
        edge[i].flow--;    edge[i].val--;
        if(to<=lim)    q[++tot]=calc(to);
        dfs(q[tot].x,q[tot].y,to);
        return;
    }
}

bool ext[maxn][maxn];

int main(){
    int e=read();
    m=read();n=read();
    Start=1;    End=n*m*2;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j){
            int x=read();
            if(i!=1&&ext[i-1][j]==0)    add(T(i-1,j),F(i,j),0x7fffffff,0);
            if(j!=1&&ext[i][j-1]==0)    add(T(i,j-1),F(i,j),0x7fffffff,0);
            if(x==1){
                ext[i][j]=1;
                continue;
            }
            add(F(i,j),T(i,j),1,x==2?1:0);
            add(F(i,j),T(i,j),0x7fffffff,0);

        }
    int cnt=0;
    while(1){
        int now=spfa();
        if(now<0)    break;
        cnt++;
        if(cnt>e)    break;
        tot=0;
        dfs(1,1,1);
        q[0]=(Node){1,1};
        for(int j=1;j<=tot;++j){
            Node a=q[j-1],b=q[j];
            if(a.x==b.x)    printf("%d %d\n",cnt,1);
            else            printf("%d %d\n",cnt,0);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/cellular-automaton/p/8449320.html