HihoCoder 1424Asa's Chess Problem

来源：互联网发布：java页面跳转代码编辑：程序博客网时间：2024/06/05 19:15

一开始搜上下界没有见到费用流，就以为考不到，没有学。今天接着碰到了。。。
其实跟有源汇上下界最大流差不多。也是带了一些有花费的边。
对于这个题，一开始也没分析好。对于每行和每列的点，跟s相连的是已经有的黑子的数目，这是必须流的，也就是下界，上界很明显就是相同的。
然后跟e相连的是题目要求的上下界流量。然后交换怎么体现呢？题目中很亲民的给出每一对点要么行相同，要么列相同，很明显行相同，只会对列有影响，反之，只会对行有影响，那么我们就加上这样的一条边，对于不同的行或列连边，花费为1，下界0，上界1。

#include <cstdio>#include <queue>#include <cstring>#include <algorithm>#include <cstring>using namespace std;const int MAXN = 3000+5;const int inf = 1e9;typedef long long LL;int n;int sr[52],sc[52];int f[MAXN];int cnt,head[MAXN];int max_flow,min_cost,sum_flow;struct node{    int u,v,w,f,next;} edge[100000];void init(){    cnt = 0;    memset(head,-1,sizeof head);    memset(f,0,sizeof f);    memset(sr,0,sizeof sr);    memset(sc,0,sizeof sc);}void add(int u,int v,int w,int f){    edge[cnt].u = u;    edge[cnt].v = v;    edge[cnt].w = w;    edge[cnt].f = f;    edge[cnt].next = head[u];    head[u] = cnt++;    edge[cnt].u = v;    edge[cnt].v = u;    edge[cnt].w = -w;    edge[cnt].f = 0;    edge[cnt].next = head[v];    head[v] = cnt++;}void Add(int u,int v,int w,int fl,int fh){    f[u] -= fl;    f[v] += fl;    add(u,v,w,fh-fl);}bool vis[MAXN];int dis[MAXN],pre[MAXN];bool spfa(int s,int e){    for(int i = 0; i <= e; ++i)    {        dis[i] = inf;        pre[i] = -1;    }    queue<int>q;    q.push(s);    dis[s] = 0;    while(!q.empty())    {        int u = q.front();        q.pop();        vis[u] = 0;        for(int i = head[u]; i != -1; i = edge[i].next)        {            int v = edge[i].v;            int w = edge[i].w;            int f = edge[i].f;            if(f > 0 && dis[v] > dis[u] + w)            {                dis[v] = dis[u] + w;                pre[v] = i;                if(!vis[v])                {                    vis[v] = 1;                    q.push(v);                }            }        }    }    if(pre[e] == -1)return 0;    return 1;}void get_mincost(int s,int e){    max_flow = 0,min_cost = 0;    while(spfa(s,e))    {        int p = pre[e];        int flow = inf;        while(p != -1)        {            flow = min(flow,edge[p].f);            p = pre[edge[p].u];        }        max_flow += flow;        min_cost += flow*dis[e];        p = pre[e];        while(p != -1)        {            edge[p].f -= flow;            edge[p^1].f += flow;            p = pre[edge[p].u];        }    }    if(max_flow == sum_flow)printf("%d\n",min_cost);    else puts("-1");}int tu[52][52];int main(){    while(~scanf("%d",&n))    {        init();        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= n; ++j)            {                scanf("%d",&tu[i][j]);                if(tu[i][j])sr[i]++,sc[j]++;            }        int s = 0,e = 2*n+1;        for(int i = 1; i <= n; ++i)Add(s,i,0,sr[i],sr[i]);        for(int i = 1; i <= n; ++i)Add(s,i+n,0,sc[i],sc[i]);        int l,r;        for(int i = 1; i <= n; ++i)        {            scanf("%d%d",&l,&r);            Add(i,e,0,l,r);        }        for(int i = 1; i <= n; ++i)        {            scanf("%d%d",&l,&r);            Add(i+n,e,0,l,r);        }        int u,v,u1,v1;        for(int i = 0,R = n*n/2; i < R; ++i)        {            scanf("%d%d%d%d",&u,&v,&u1,&v1);            if(tu[u][v] && !tu[u1][v1])            {                if(u == u1)                {                    Add(v+n,v1+n,1,0,1);                }                else Add(u,u1,1,0,1);            }            else if(!tu[u][v] && tu[u1][v1])            {                if(u == u1)                {                    Add(v1+n,v+n,1,0,1);                }                else Add(u1,u,1,0,1);            }        }        int ss = e+1,ee = e+2;        sum_flow = 0;        for(int i = 0; i <= e; ++i)        {            if(f[i] > 0)            {                add(ss,i,0,f[i]);                sum_flow += f[i];            }            else add(i,ee,0,-f[i]);        }        add(e,s,0,inf);        get_mincost(ss,ee);    }    return 0;}

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