bzoj5251: [2018多省省队联测]劈配

题目链接

bzoj5251: [2018多省省队联测]劈配

题解

在上一个人匹配完图的基础上，我们可以对每个人的志愿分别匈牙利算法二分图匹配
匹配时,从小到大枚举该人的志愿匹配，复杂度是\(O(nC+m)\)，然后第一问做完了
第二问\(~~我们可以二分答案，匈牙利算法~~\)我们暴力枚举在i位置之后的每个位置k,做一次假的二分图匹配只寻找，不修改...判断能否匹配成功
然后我们需要匈牙利\(n^2\)次
所以总复杂度\(O(n^3C+n^2m)\)
ε=(´ο｀*)))唉，考场上还是太Naive了QAQQQQQQ

代码

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using std::vector;
inline int read() {
    int x = 0,f = 1;char c = getchar();
    while(c < '0' ||  c > '9') {if(c == '-') f = -1;c = getchar();} 
    while(c <= '9' && c >= '0') x=x * 10 +c - '0' ,c = getchar();
    return x *f ;
}/*
inline void add_edge(int u ,int v,int flow ,int cost) {
    edge[++num].v = v;edge[num].flow = flow;edge[num].next =head[x];haed[x] = num;
}*/
const int maxn = 207;
int T,C,n,m; 
int a[maxn][maxn],b[maxn];
vector<int>vec[maxn][maxn],que[maxn];  
int s[maxn],ans[maxn],match[maxn],Wmatch[maxn];  
bool vis[maxn]; 
void init() { 
    for(int i = 1;i <= n;++ i) que[i].clear();  
        for(int i = 1;i <= n;++ i) for(int j = 1;j <= m;++ j) vec[i][j].clear(),
        s[i] = b[i] = match[i] = Wmatch[i] = ans[i] = 0 ;
    n = read(), m = read();
        for(int i = 1;i <= m;++ i) b[i] = read(),que[i].clear(); 
        for(int i = 1;i <= n;++ i) 
            for(int a,j = 1;j <= m;++ j) a = read(),vec[i][a].push_back(j) ;  
        for(int i = 1;i <= n;++ i) s[i] = read(); 
} 
 
bool dfs(int x,bool flag) { 
    int l,r; if(!flag)l = Wmatch[x] == 0 ? 1 : Wmatch[x],r = Wmatch[x] == 0 ? s[x] : Wmatch[x]; 
         else l = Wmatch[x] == 0 ? 1 : Wmatch[x],r = Wmatch[x] == 0 ? m : Wmatch[x]; 
    for(int i = l;i <= r;++ i) { 
        for(int j = 0;j < vec[x][i].size(); ++ j) {
            int v = vec[x][i][j]; 
            if(vis[v]) continue;
            vis[v] = 1; 
            //printf("%d\n",que[v].size()); 
            if(que[v].size() < b[v]) { 
                if(flag) {
                    que[v].push_back(x); 
                    match[x] = v;
                    Wmatch[x] = i; 
                } 
                return true; 
            }
            for(int k = 0;k < que[v].size();k ++ ) {
                int tv = que[v][k]; 
                if(dfs(tv,flag)) { 
                    if(flag) { 
                        que[v].erase(que[v].begin() + k); 
                        match[x] = v; 
                        Wmatch[x] = i; 
                        que[v].push_back(x); 
                    } 
                    return true; 
                } 
                 
            }
             
            //if(que[2].size() == 1) puts("!!!!!!!!!!!!");  
        } 
    } 
    return false; 
} 
int main() { 
        T = read();C = read(); 
        while(T --) { 
            init(); 
        for(int i = 1;i <= n;++ i) { 
            for(int j = i;j <= n;++ j) { 
                memset(vis,0,sizeof vis); 
                if(dfs(j,false)) ans[j] = i;   
            } 
            memset(vis,false,sizeof vis); 
                dfs(i,true);    
            printf("%d ",Wmatch[i] ? Wmatch[i] : m + 1);  
        } 
        puts(""); 
        for(int i = 1;i <= n; ++ i) { 
            printf("%d ",i - ans[i]); 
        } 
        puts(""); 
        } 
        return 0;
}