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  • 【Luogu】P2445动物园(最大流)

      题目链接

      题目本身还是比较水的吧……容易发现是最大流套上dinic跑一遍就好了,并不会超时。

      比较不偷税的一点是关于某动物的所有目击报告都符合才能连边……qwqqwqqwq

      

    #include<cstdio>
    #include<algorithm>
    #include<cctype>
    #include<cstring>
    #include<cstdlib>
    #include<queue>
    #define maxn 1000
    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;
    }
    
    int u[5]={0,0,1,0,-1};
    int w[5]={0,1,0,-1,0};
    
    struct Edge{
        int next,to,val;
    }edge[maxn*2];
    int head[maxn],num;
    inline void addedge(int from,int to,int val){
        edge[++num]=(Edge){head[from],to,val};
        head[from]=num;
    }
    inline void add(int from,int to,int val){
        addedge(from,to,val);
        addedge(to,from,0);
    }
    
    inline int count(int i){    return i&1?i+1:i-1;    }
    
    int pre[maxn];
    int Start,End;
    bool vis[maxn];
    int dfn[maxn];
    int list[maxn];
    
    bool bfs(){
        memset(vis,0,sizeof(vis));    dfn[Start]=1;    vis[Start]=1;
        queue<int>q;    q.push(Start);
        while(!q.empty()){
            int from=q.front();    q.pop();
            for(int i=head[from];i;i=edge[i].next){
                int to=edge[i].to;
                if(vis[to]||edge[i].val<=0)    continue;
                vis[to]=1;    dfn[to]=dfn[from]+1;
                q.push(to);
            }
        }
        return vis[End];
    }
    
    int dfs(int x,int val){
        if(val==0||x==End)    return val;
        vis[x]=1;    int flow=0;
        for(int &i=list[x];i;i=edge[i].next){
            int to=edge[i].to;
            if(vis[to]||dfn[to]!=dfn[x]+1||edge[i].val<=0)    continue;
            int now=dfs(to,min(val,edge[i].val));
            if(x!=End)    pre[to]=x;    edge[i].val-=now;    edge[count(i)].val+=now;    val-=now;    flow+=now;
            if(val<=0)    break;
        }
        if(val!=flow)    dfn[x]=-1;
        return flow;
    }
    
    inline int maxflow(){
        int ans=0;
        while(bfs()){
            memset(vis,0,sizeof(vis));
            for(int i=Start;i<=End;++i)    list[i]=head[i];
            int now=dfs(Start,0x7fffffff);
            if(now==0)    break;
            ans+=now;
        }
        return ans;
    }
    
    int s[maxn][maxn];
    char c[maxn];
    bool ext[maxn][maxn];
    int dis[105][105][105];
    int v[maxn];
    bool exa[maxn];
    int f[maxn][maxn];
    int sum[maxn];
    
    struct Node{
        int x,y;
    }que[maxn];
    
    int main(){
        memset(dis,127/3,sizeof(dis));    int inf=dis[0][0][0];
        int n=read();
        for(int i=1;i<=n;++i){
            scanf("%s",c+1);
            for(int j=1;j<=n;++j)
                if(c[j]=='*')    s[i][j]=1;
                else             s[i][j]=0;
        }
        int p=read();    End=p*2+1;
        for(int i=1;i<=p;++i)    que[i]=(Node){read(),read()};
        for(int i=1;i<=p;++i){
            add(Start,i,1);    add(i+p,End,1);
            queue<Node>q;    q.push(que[i]);    dis[i][que[i].x][que[i].y]=0;
            while(!q.empty()){
                Node from=q.front();    q.pop();
                //printf("%d %d %d<<<<<<<<
    ",i,from.x,from.y);
                for(int j=1;j<5;++j){
                    int nx=from.x+u[j];
                    int ny=from.y+w[j];
                    //printf("%d %d %d %d><>
    ",nx,ny,s[nx][ny],dis[i][from.x][from.y]);
                    if(nx<1||nx>n||ny<1||ny>n||s[nx][ny]||dis[i][nx][ny]!=inf)    continue;
                    dis[i][nx][ny]=dis[i][from.x][from.y]+1;
                    q.push((Node){nx,ny});
                }
            }
        }
        for(int i=1;i<=p;++i)    v[i]=read();
        int r=read();
        for(int i=1;i<=r;++i){
            int t=read(),x=read(),y=read(),d=read();    sum[d]++;
            for(int j=1;j<=p;++j){
                //printf("%d %d>>>
    ",j,dis[j][x][y]);
                int ret=v[d]*t;
                if(ret>=dis[j][x][y])    f[j][d]++;
            }
        }
        for(int i=1;i<=p;++i)
            for(int j=1;j<=p;++j)
                if(sum[i]==f[j][i])    add(j,i+p,1);
        maxflow();
        for(int i=1+p;i<End;++i){
            for(int j=head[i];j;j=edge[j].next){
                int to=edge[j].to;
                if(to==End||edge[j].val==0)    continue;
                printf("%d %d %d
    ",i-p,que[to].x,que[to].y);
                break;
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cellular-automaton/p/8820184.html
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