【BZOJ 1305】[CQOI2009]dance跳舞

【链接】 我是链接,点我呀:)

【题意】

在这里输入题意

【题解】

男生和女生每个人都分身成两个节点
即x[1],x[2]和y[1],y[2]

然后如果i和j不相互喜欢

那么add(x[i][2],y[j][2],1)

如果相互喜欢的话

add(x[i][1],y[j][1],1)

然后对于每个男生

add(x[i][1],y[i][1],k)

对于每个女生

add(y[i][2],y[i][1],k)

然后对于每个男生

add(s,x[i][1],mid)

add(y[i][1],t,mid)

这里的mid是二分的值。

这个mid就是舞会的轮数了。

如果能够满流显然每个人都能配对跳mid支舞

显然是有单调性的。

(然后发现原来的dicnic的模板是错的。。边数的计算要特别careful.

【代码】

#include <bits/stdc++.h>
using namespace std;

const int N = 50;

struct abc{
    int nex,en,flow;
}bian[2][N*N*2+8*N+10];

int n,k,x[N+10][3],y[N+10][3],fir[2][N*4+10],tfir[N*4+10],totm,deep[N*4+10];
int cnt = 0;
bool bo[N+10][N+10];
char s[N+10];
queue<int> dl;

//每个男人分为(x1,x2)
//每个女人分为(y1,y2)

void add(int x,int y,int cost,int i){
    bian[i][totm].nex = fir[i][x];
    fir[i][x] = totm;
    bian[i][totm].en = y,bian[i][totm].flow = cost;
    totm++;

    bian[i][totm].nex = fir[i][y];
    fir[i][y] = totm;
    bian[i][totm].en = x,bian[i][totm].flow = 0;
    totm++;
}

bool bfs(int s,int t){
    dl.push(s);
    memset(deep,255,sizeof deep);
    deep[s] = 0;

    while (!dl.empty()){
        int x = dl.front();
        dl.pop();
        for (int temp = fir[1][x]; temp!= -1 ;temp = bian[1][temp].nex){
            int y = bian[1][temp].en;
            if (deep[y]==-1 && bian[1][temp].flow>0){
                deep[y] = deep[x] + 1;
                dl.push(y);
            }
        }
    }
    return deep[t]!=-1;
}

int dfs(int x,int t,int limit){
    if (x == t) return limit;
    if (limit == 0) return 0;
    int cur,f = 0;
    for (int temp = tfir[x];temp!=-1;temp = bian[1][temp].nex){
        tfir[x] = temp;
        int y = bian[1][temp].en;
        if (deep[y] == deep[x] + 1 && (cur = dfs(y,t,min(limit,bian[1][temp].flow))) ){
            f += cur;
            limit -= cur;
            bian[1][temp].flow -= cur;
            bian[1][temp^1].flow += cur;
            if (!limit) break;
        }
    }
    return f;
}

int get_flow(){
    int now = 0;
    while (bfs(0,cnt)){
        for (int i = 0;i <= cnt;i++) tfir[i] = fir[1][i];
        int xxx = dfs(0,cnt,10000);
        now+=xxx;
    }
    return now;
}

int main(){
    //freopen("F:\\program\\rush\\rush_in.txt","r",stdin);
    ios::sync_with_stdio(0),cin.tie(0);
    memset(fir[0],255,sizeof fir[0]);
    cin >> n >> k;
    for (int i = 1;i <= n;i++){
        cin >> (s+1);
        for (int j = 1;j <= n;j++)
            if (s[j]=='Y'){
                bo[i][j] = 1;
            }
    }
    for (int i = 1;i <= n;i++){
        for (int j = 1;j <= 2;j++)
            x[i][j] = ++cnt;
    }

    for (int i = 1;i <= n;i++)
        for (int j = 1;j <= 2;j++)
            y[i][j] = ++cnt;

    cnt++;
    for (int i = 1;i <= n;i++){
        for (int j = 1;j <= n;j++){
            if (bo[i][j])
                add(x[i][1],y[j][1],1,0);
            else{
                add(x[i][2],y[j][2],1,0);
            }
        }
    }

    for (int i = 1;i <= n;i++){
        add(x[i][1],x[i][2],k,0);
        add(y[i][2],y[i][1],k,0);
    }

    int l = 0,r = n+1,temp1 = 0;
    while (l<=r){
        int i = (l+r)>>1;

        for (int j = 0;j < totm;j++) bian[1][j] = bian[0][j];
        for (int j = 0;j <= cnt;j++) fir[1][j] = fir[0][j];
        int tt = totm;

        for (int j = 1;j <= n;j++){
            add(0,x[j][1],i,1);
            add(y[j][1],cnt,i,1);
        }
        int temp = get_flow();

        if (temp!=i*n){
            r = i-1;
        }else {
            temp1 = i;
            l = i+1;
        }
        totm = tt;
    }
    cout<<temp1<<endl;
    return 0;
}