HDU 2119 Matrix 最小点覆盖

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题目：

http://acm.hdu.edu.cn/showproblem.php?pid=2119

题意：

给定一个n*m的矩阵，里面的元素只有0和1，每次可以选择一行或者一列，把此行或者此列的1全部变成0，问最少几次可以使矩阵元素只有0

思路：

把行看成一个点集，列看成一个点集，于是就是一个二分图，第i行j列为1，便在两点之间连一条边，于是题目就变成了二分图的最小点覆盖，用匈牙利算法或者最大流可解
最大流：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <set>#include <cmath>using namespace std;typedef long long ll;const int N = 210, INF = 0x3f3f3f3f;struct edge{    int to, cap, next;} g[N*N*2];int cnt, head[N];int gap[N], que[N], level[N], pre[N], cur[N];int ss, tt, nv;void add_edge(int v, int u, int cap){    g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;    g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;}void bfs(int t){    memset(level, -1, sizeof level);    memset(gap, 0, sizeof gap);    int st = 0, en = 0;    level[t] = 0;    que[en++] = t;    gap[level[t]]++;    while(st < en)    {        int v = que[st++];        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(level[u] < 0)            {                level[u] = level[v] + 1;                gap[level[u]]++;                que[en++] = u;            }        }    }}int sap(int s, int t){    bfs(t);    memcpy(cur, head, sizeof head);    int v = pre[s] = s, flow = 0, aug = INF;    while(level[s] < nv)    {        bool flag = false;        for(int &i = cur[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && level[v] == level[u] + 1)            {                flag = true;                pre[u] = v;                v = u;                aug = min(aug, g[i].cap);                if(v == t)                {                    flow += aug;                    while(v != s)                    {                        v = pre[v];                        g[cur[v]].cap -= aug;                        g[cur[v]^1].cap += aug;                    }                    aug = INF;                }                break;            }        }        if(flag) continue;        int minlevel = nv;        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && level[u] < minlevel)                minlevel = level[u], cur[v] = i;        }        if(--gap[level[v]] == 0) break;        level[v] = minlevel + 1;        gap[level[v]]++;        v = pre[v];    }    return flow;}int main(){    int n, m, a;    while(scanf("%d", &n), n)    {        scanf("%d", &m);        cnt = 0;        memset(head, -1, sizeof head);        ss = 0, tt = n + m + 1;        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)            {                scanf("%d", &a);                if(a) add_edge(i, n + j, 1);            }        for(int i = 1; i <= n; i++) add_edge(ss, i, 1);        for(int i = 1; i <= m; i++) add_edge(n + i, tt, 1);        nv = tt + 1;        printf("%d\n", sap(ss, tt));    }    return 0;}

匈牙利算法：

#include <bits/stdc++.h>using namespace std;const int N = 210;struct edge{    int to, next;} g[N*N*2];int head[N], match[N];int nx, ny, cnt, cas = 0;bool used[N];void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}bool dfs(int v){    for(int i = head[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(! used[u])        {            used[u] = true;            if(match[u] == -1 || dfs(match[u]))            {                match[u] = v;                return true;            }        }    }    return false;}int hungary(){    int res = 0;    memset(match, -1, sizeof match);    for(int i = 1; i <= nx; i++)    {        memset(used, 0, sizeof used);        if(dfs(i)) res++;    }    return res;}int main(){    int n, m, a;    while(scanf("%d", &n), n)    {        scanf("%d", &m);        cnt = 0;        memset(head, -1, sizeof head);        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)            {                scanf("%d", &a);                if(a) add_edge(i, n + j);            }        nx = n, ny = m;        printf("%d\n", hungary());    }    return 0;}

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