LightOJ 1330 Binary Matrix 最大流

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题目：http://www.lightoj.com/volume_showproblem.php?problem=1330

题意：假定有n * m的矩阵，矩阵中元素只有0和1，现在给定矩阵的每一行的和与每一列的和，复原出这个矩阵，有多个的话输出字典序最小的那个

思路：首先判断行和与列和是否相等，不相等的话肯定不可能存在满足条件的矩阵。然后把行和列作为点，从源点向所有行连边，容量为行和，从每一列向汇点连边，容量为列和，跑一次最大流，判断流量是否与行和或列和相等，不等的话也不存在满足条件的矩阵。发现有满足条件的矩阵，如何取字典序最小呢，此时从源点ss到行、从列到汇点tt容量都为0，假设(i, j)位置上为1，首先我们把i -> j之间正向弧和反向弧的容量都清空（无论（i，j）位置是不是1,），然后从ss -> i, j -> tt容量均设为1，如果能跑出流量，说明可以把当前位置的1向后移动，那么当前位置就可以是0，若不能跑出流量就不能向后移动，至于跑出流量就说明当前位置的1可以向后移动的原因，是因为此点之前的容量都被清空了，不可能从前面流过，只能从此点之后流过。此题dinic比sap快了一倍。。。

sap算法

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#define debug() puts("here");using namespace std;const int N = 110;const int INF = 0x3f3f3f3f;struct edge{    int to, cap, id, next;} g[N*N*2];int cnt, nv, head[N], level[N], gap[N], cur[N], pre[N];int id[N][N];int cas;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++;}int sap(int s, int t){    memset(level, 0, sizeof level);    memset(gap, 0, sizeof gap);    memcpy(cur, head, sizeof head);    gap[0] = nv;    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 && minlevel > level[u])                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 t, n, m, a;    scanf("%d", &t);    while(t--)    {        cnt = 0;        memset(head, -1, sizeof head);        memset(id, 0, sizeof id);        scanf("%d%d", &n, &m);        int ss = 0, tt = n + m + 1;        int sum1 = 0, sum2 = 0;        for(int i = 1; i <= n; i++) scanf("%d", &a), add_edge(ss, i, a), sum1 += a, id[ss][i] = cnt - 1;        for(int i = 1; i <= m; i++) scanf("%d", &a), add_edge(n + i, tt, a), sum2 += a, id[n+i][tt] = cnt - 1;        if(sum1 != sum2)        {            printf("Case %d: impossible\n", ++cas); continue;        }        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)                add_edge(i, n + j, 1), id[i][n+j] = cnt - 1;        nv = tt + 1;        int res = sap(ss, tt);        if(res != sum1)        {            printf("Case %d: impossible\n", ++cas); continue;        }        printf("Case %d:\n", ++cas);        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)            {                int f = g[id[i][n+j]].cap;                g[id[i][n+j]].cap = g[id[i][n+j]^1].cap = 0;                if(f)                {                    g[id[ss][i]^1].cap = g[id[n+j][tt]^1].cap = 1;                    if(sap(ss, tt)) f = 0;                    else g[id[ss][i]^1].cap = g[id[n+j][tt]^1].cap = 0;                }                printf("%d", f);                if(j == m) printf("\n");            }    }    return 0;}

dinic算法

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#define debug() puts("here");using namespace std; const int N = 110;const int INF = 0x3f3f3f3f;struct edge{    int to, cap, id, next;} g[N*N*2];int head[N], iter[N], level[N];int id[N][N];int n, m, cnt;int cas;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++;}bool bfs(int s, int t){    memset(level, -1, sizeof level);    level[s] = 0;    queue<int> que;    que.push(s);    while(! que.empty())    {        int v = que.front(); que.pop();        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && level[u] < 0)            {                level[u] = level[v] + 1;                que.push(u);            }        }    }    return level[t] == -1;}int dfs(int v, int t, int f){    if(v == t) return f;    for(int &i = iter[v]; i != -1; i = g[i].next)    {        int u = g[i].to;        if(g[i].cap > 0 && level[v] < level[u])        {            int d = dfs(u, t, min(g[i].cap, f));            if(d > 0)            {                g[i].cap -= d, g[i^1].cap += d;                return d;            }        }    }    return 0;}int dinic(int s, int t){    int flow = 0, f;    while(true)    {        if(bfs(s, t)) return flow;        memcpy(iter, head, sizeof head);        while(f = dfs(s, t, INF),f > 0)            flow += f;    }}int main(){    int t, n, m, a;    scanf("%d", &t);    while(t--)    {        cnt = 0;        memset(head, -1, sizeof head);        memset(id, 0, sizeof id);        scanf("%d%d", &n, &m);        int ss = 0, tt = n + m + 1;        int sum1 = 0, sum2 = 0;        for(int i = 1; i <= n; i++) scanf("%d", &a), add_edge(ss, i, a), sum1 += a, id[ss][i] = cnt - 1;        for(int i = 1; i <= m; i++) scanf("%d", &a), add_edge(n + i, tt, a), sum2 += a, id[n+i][tt] = cnt - 1;        if(sum1 != sum2)        {            printf("Case %d: impossible\n", ++cas); continue;        }        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)                add_edge(i, n + j, 1), id[i][n+j] = cnt - 1;        int res = dinic(ss, tt);        if(res != sum1)        {            printf("Case %d: impossible\n", ++cas); continue;        }        printf("Case %d:\n", ++cas);        for(int i = 1; i <= n; i++)            for(int j = 1; j <= m; j++)            {                int f = g[id[i][n+j]].cap;                g[id[i][n+j]].cap = g[id[i][n+j]^1].cap = 0;                if(f)                {                    g[id[ss][i]^1].cap = g[id[n+j][tt]^1].cap = 1;                    if(dinic(ss, tt)) f = 0;                    else g[id[ss][i]^1].cap = g[id[n+j][tt]^1].cap = 0;                }                printf("%d", f);                if(j == m) printf("\n");            }    }    return 0;}

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