最大流+二分 bzoj1189 緊急疏散__bzoj

題目連結：http://www.lydsy.com/JudgeOnline/problem.php?id=1189
題目大意：給你一個n*m的地區，有門和牆，每塊空地上有一個人，疏散開始後每塊空地可以站無數個人，但是門每次只能經過一個人，請問最少需要多少時間才能全部撤到門外

建一個源點s串連到所有空地，流量為1，然後二分最長時間，空地向可以到達的門連一條邊(這裡的可到達是指在二分出的時間內可到達)，所有門向匯點t連一條容量為時間的邊，跑最大流，如果滿流，就降低上限，否則提高下限。(似乎直接從1暴力跑下去也不會TLE……)

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <cstring>#include <cstdlib>#define INF 214748364using namespace std;struct Link{    int s,t,c,next;}l[100000];Link l2[100000];int g[5000];int cnt[5000];int dist[5000];bool b[5000];char a[30][30];int d[30][30];queue<int> Q;int n,m,p,sum = 0;bool bfs(int x,int y){    Q.push(x);    dist[x] = 1;    while(!Q.empty())    {        int q = Q.front();        int w = g[q];        while(w)        {            if(l[w].c && !dist[l[w].t])            {                dist[l[w].t] = dist[q] + 1;                Q.push(l[w].t);            }            w = l[w].next;        }        Q.pop();    }    return dist[y];}int dfs(int x,int t,int flow){    if(x == t || flow == 0)        return flow;    int w = cnt[x];    while(w)    {        if(dist[l[w].t] == dist[x] + 1 && !b[l[w].t] && l[w].c)        {            b[l[w].t] = 1;            int k = dfs(l[w].t,t,min(flow,l[w].c));            b[l[w].t] = 0;            if(k)            {                cnt[x] = w;                l[w].c -= k;                l[w^1].c += k;                return k;            }        }        w = l[w].next;    }    return 0;}int maxflow(int s,int t){    memset(dist,0,sizeof(dist));    int flow = 0;    while(bfs(s,t))    {        while(233)        {            int w = dfs(s,t,INF);            memset(b,0,sizeof(b));            flow += w;            if(w == 0)                break;        }        for(int i = 0;i <= t;i ++)            cnt[i] = g[i];        memset(dist,0,sizeof(dist));    }    return flow;}void Add_Link(int s,int t,int c,int x){    l[x].s = s;    l2[x].s = l[x].s;    l[x].t = t;    l2[x].t = l[x].t;    l[x].c = c;    l2[x].c = l[x].c;    l[x].next = g[s];    g[s] = x;    l[x^1].s = t;    l2[x^1].s = l[x^1].s;    l[x^1].t = s;    l2[x^1].t = l[x^1].t;    l[x^1].c = 0;    l2[x^1].c = l[x^1].c;    l[x^1].next = g[t];    g[t] = x^1;    return ;}void bbffss(int x,int y,int time){    int k = (x-1)*(m+1) + y;    Q.push(k);    while(!Q.empty())    {        int e = Q.front();        x = e / (m+1) + 1;        y = e % (m+1);        p += 2;        Add_Link((x-1)*(m+1)+y,k,1,p);        if(!d[x+1][y] && a[x+1][y] == '.' )        {            d[x+1][y] = d[x][y] + 1;            if(d[x+1][y] <= time)                Q.push((x)*(m+1)+y);        }        if(!d[x-1][y] && a[x-1][y] == '.')        {            d[x-1][y] = d[x][y] + 1;            if(d[x-1][y] <= time)                Q.push((x-2)*(m+1)+y);        }        if(!d[x][y+1] && a[x][y+1] == '.')        {            d[x][y+1] = d[x][y] + 1;            if(d[x][y+1] <= time)                Q.push((x-1)*(m+1)+y+1);        }        if(!d[x][y-1] && a[x][y-1] == '.')        {            d[x][y-1] = d[x][y] + 1;            if(d[x+1][y] <= time)                Q.push((x-1)*(m+1)+y-1);        }        Q.pop();    }    for(int i = 1;i <= n;i ++)    {        for(int j = 1;j <= m;j ++)        {            d[i][j] = 0;        }    }    return ;}int main(){    scanf("%d%d",&n,&m);    for(int i = 1;i <= n;i ++)    {        cin >> a[i];        for(int j = m;j ;j --)        {            a[i][j] = a[i][j-1];            if(a[i][j] == '.')            {                p += 2;                Add_Link(0,(i-1)*(m+1)+j,1,p);                sum ++;            }        }        a[i][0] = 'X';        a[i][m+1] = 'X';    }    for(int i = 1;i <= m;i ++)    {        a[0][i] = 'X';        a[n+1][i] = 'X';    }    int s = 1,t = 1000,k = p + 2;    while(s != t)    {        int mid = (s+t)/2;        p = 0;        memset(l,0,sizeof(l));        memset(g,0,sizeof(g));        for(p ;p < k;p += 2)            Add_Link(l2[p].s,l2[p].t,l2[p].c,p);        for(int i = 1;i <= n;i ++)        {            for(int j = 1;j <= m;j ++)            {                if(a[i][j] == 'D')                {                    bbffss(i,j,mid);                    p += 2;                    Add_Link((i-1)*(m+1)+j,4999,mid,p);                }            }        }        if(maxflow(0,4999) == sum)            t = mid;        else            s = mid + 1;    }    if(s == 1000)    {        cout << "impossible" << endl;        return 0;    }    cout << s << endl;    return 0;}