标签：printf \n return max sharp queue 转化 from algorithm
http://fastvj.rainng.com/contest/236779#problem/G
Description：
n 行 m 列
给你行和 与 列和
然后有Q个限制，表示特定单元格元素大小的范围，最后问你可行的矩阵值
Solution：
有源汇上下界最大流问题，初始源点 连 行和流量是该行对应得行和，然后列连初始汇点，容量为列的列和，详细的限制，对应于行列之间，因为我最后要输出一个矩阵，所以n * m每一条边都要链接最后，手动连一条 t s inf的边，变成无源汇有上下界可行流问题，然后拆边，把对应边的流量，放到数组里，输出就好啦
Code：
前面比较基础的Dinic，数据操作函数，加边操作
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define inf (1 << 28)
using namespace std;
const int maxn = 25;
const int maxm = 210;
const int mn = 505;
const int mm = 440020;
struct node{
int to,val,pre,lid;
}e[mm];
int id[mn],cnt=0;
int cur[mn];
int flor[mn];
int upflow[mn];
int lowf[maxm][maxn];
int upf[maxm][maxn];
int out[maxm][maxn];
void init()
{
memset(id,-1,sizeof(id));
memset(upflow,0,sizeof(upflow));
cnt = 0;
}
void add(int from,int to,int val,int lid)
{
e[cnt].to = to;
e[cnt].val = val;
e[cnt].lid = lid;
e[cnt].pre = id[from];
id[from] = cnt++;
swap(from,to);
e[cnt].to = to;
e[cnt].val = 0;
e[cnt].lid = lid;
e[cnt].pre = id[from];
id[from] = cnt++;
}
void addflow(int from,int to,int low,int up,int lid)
{
upflow[from] -= low;
upflow[to] += low;
add(from,to,up-low,lid);
}
bool bfs(int s,int t)
{
memset(flor,0,sizeof(flor));
flor[s] = 1;
queue<int> q;
while(q.size())q.pop();
q.push(s);
while(q.size())
{
int now = q.front();
q.pop();
for(int i = id[now];~i;i = e[i].pre)
{
int to = e[i].to;
int val = e[i].val;
if(val > 0 && flor[to] == 0)
{
flor[to] = flor[now] + 1;
q.push(to);
if(to == t)
return true;
}
}
}
return false;
}
int dfs(int s,int t,int value)
{
if(s == t || value == 0)return value;
int ret = value,a;
for(int &i = cur[s];~i;i = e[i].pre)
{
int val = e[i].val;
int to = e[i].to;
if(flor[to] == flor[s] + 1 && (a = dfs(to,t,min(ret,val))))
{
e[i].val -= a;
e[i^1].val += a;
ret -= a;
if(ret == 0)break;
}
}
if(ret == value)flor[s] = 0;
return value - ret;
}
int Dinic(int s,int t)
{
int ret = 0;
while(bfs(s,t))
{
memcpy(cur,id,sizeof(id));
ret += dfs(s,t,inf);
}
return ret;
}
void addlimit(int i,int j,char op,int lim)
{
if(op == ‘=‘)
{
upf[i][j] = lowf[i][j] = lim;
}
else if(op == ‘>‘)
{
lowf[i][j] = max(lowf[i][j],lim+1);
}
else
{
upf[i][j] = min(upf[i][j],lim-1);
}
}
然后根据输入一点点的加边，这是行和和列和对应的边
scanf("%d%d",&n,&m);
s = 0;
t = n + m + 1;
ss = t + 1;
tt = ss + 1;
int lsum;
for(int i = 1;i <= n;++i)
{
scanf("%d",&lsum);
addflow(s,i,lsum,lsum,0);
}
for(int i = 1;i <= m;++i)
{
scanf("%d",&lsum);
addflow(n + i,t,lsum,lsum,0);
}
int limitnum;
然后根据题目中给出的限制条件，填充上下界数组
scanf("%d",&limitnum);
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
lowf[i][j] = 0;
upf[i][j] = inf;
}
}
int row,col,lim;
char op;
for(int i = 1;i <= limitnum;++i)
{
scanf("%d %d %c %d",&row,&col,&op,&lim);
if(row == 0 && col == 0)
{
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
addlimit(i,j,op,lim);
}
}
}
else if(row == 0)
{
for(int i = 1;i <= n;++i)
{
addlimit(i,col,op,lim);
}
}
else if(col == 0)
{
for(int i = 1;i <= m;++i)
{
addlimit(row,i,op,lim);
}
}
else
{
addlimit(row,col,op,lim);
}
}
根据上下界数组，添加有上下界边
int tot = 0;
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
addflow(i,n+j,lowf[i][j],upf[i][j],++tot);
}
}
加边完成后，转化为无源汇有上下界可行流 ———— 循环流
add(t,s,inf,0);
int sum = 0;
for(int i = s;i <= t;++i)
{
if(upflow[i] < 0)
{
add(i,tt,-upflow[i],0);
}
else
{
sum += upflow[i];
add(ss,i,upflow[i],0);
}
}
跑DIinc，如果存在可行流的话
就输出叭~~
for(int now = n+1;now <= n + m;++now)
{
for(int i = id[now];~i;i = e[i].pre)
{
int to = e[i].to;
int lid = e[i].lid;
if(lid == 0 || i % 2 == 0)continue;
out[to][now-n] = lowf[to][now-n] + e[i].val;
}
}
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
if(j == 1)
printf("%d",out[i][j]);
else
printf(" %d",out[i][j]);
}
printf("\n");
}
完整代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define inf (1 << 28)
using namespace std;
const int maxn = 25;
const int maxm = 210;
const int mn = 505;
const int mm = 440020;
struct node{
int to,val,pre,lid;
}e[mm];
int id[mn],cnt=0;
int cur[mn];
int flor[mn];
int upflow[mn];
int lowf[maxm][maxn];
int upf[maxm][maxn];
int out[maxm][maxn];
void init()
{
memset(id,-1,sizeof(id));
memset(upflow,0,sizeof(upflow));
cnt = 0;
}
void add(int from,int to,int val,int lid)
{
e[cnt].to = to;
e[cnt].val = val;
e[cnt].lid = lid;
e[cnt].pre = id[from];
id[from] = cnt++;
swap(from,to);
e[cnt].to = to;
e[cnt].val = 0;
e[cnt].lid = lid;
e[cnt].pre = id[from];
id[from] = cnt++;
}
void addflow(int from,int to,int low,int up,int lid)
{
upflow[from] -= low;
upflow[to] += low;
add(from,to,up-low,lid);
}
bool bfs(int s,int t)
{
memset(flor,0,sizeof(flor));
flor[s] = 1;
queue<int> q;
while(q.size())q.pop();
q.push(s);
while(q.size())
{
int now = q.front();
q.pop();
for(int i = id[now];~i;i = e[i].pre)
{
int to = e[i].to;
int val = e[i].val;
if(val > 0 && flor[to] == 0)
{
flor[to] = flor[now] + 1;
q.push(to);
if(to == t)
return true;
}
}
}
return false;
}
int dfs(int s,int t,int value)
{
if(s == t || value == 0)return value;
int ret = value,a;
for(int &i = cur[s];~i;i = e[i].pre)
{
int val = e[i].val;
int to = e[i].to;
if(flor[to] == flor[s] + 1 && (a = dfs(to,t,min(ret,val))))
{
e[i].val -= a;
e[i^1].val += a;
ret -= a;
if(ret == 0)break;
}
}
if(ret == value)flor[s] = 0;
return value - ret;
}
int Dinic(int s,int t)
{
int ret = 0;
while(bfs(s,t))
{
memcpy(cur,id,sizeof(id));
ret += dfs(s,t,inf);
}
return ret;
}
void addlimit(int i,int j,char op,int lim)
{
if(op == ‘=‘)
{
upf[i][j] = lowf[i][j] = lim;
}
else if(op == ‘>‘)
{
lowf[i][j] = max(lowf[i][j],lim+1);
}
else
{
upf[i][j] = min(upf[i][j],lim-1);
}
}
int main()
{
int T;
scanf("%d",&T);
int n,m,s,t,ss,tt;
while(T--)
{
init();
scanf("%d%d",&n,&m);
s = 0;
t = n + m + 1;
ss = t + 1;
tt = ss + 1;
int lsum;
for(int i = 1;i <= n;++i)
{
scanf("%d",&lsum);
addflow(s,i,lsum,lsum,0);
}
for(int i = 1;i <= m;++i)
{
scanf("%d",&lsum);
addflow(n + i,t,lsum,lsum,0);
}
int limitnum;
scanf("%d",&limitnum);
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
lowf[i][j] = 0;
upf[i][j] = inf;
}
}
int row,col,lim;
char op;
for(int i = 1;i <= limitnum;++i)
{
scanf("%d %d %c %d",&row,&col,&op,&lim);
if(row == 0 && col == 0)
{
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
addlimit(i,j,op,lim);
}
}
}
else if(row == 0)
{
for(int i = 1;i <= n;++i)
{
addlimit(i,col,op,lim);
}
}
else if(col == 0)
{
for(int i = 1;i <= m;++i)
{
addlimit(row,i,op,lim);
}
}
else
{
addlimit(row,col,op,lim);
}
}
int tot = 0;
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
addflow(i,n+j,lowf[i][j],upf[i][j],++tot);
}
}
add(t,s,inf,0);
int sum = 0;
for(int i = s;i <= t;++i)
{
if(upflow[i] < 0)
{
add(i,tt,-upflow[i],0);
}
else
{
sum += upflow[i];
add(ss,i,upflow[i],0);
}
}
if(Dinic(ss,tt) == sum)
{
for(int now = n+1;now <= n + m;++now)
{
for(int i = id[now];~i;i = e[i].pre)
{
int to = e[i].to;
int lid = e[i].lid;
if(lid == 0 || i % 2 == 0)continue;
out[to][now-n] = lowf[to][now-n] + e[i].val;
}
}
for(int i = 1;i <= n;++i)
{
for(int j = 1;j <= m;++j)
{
if(j == 1)
printf("%d",out[i][j]);
else
printf(" %d",out[i][j]);
}
printf("\n");
}
}
else
{
printf("IMPOSSIBLE\n");
}
if(T)printf("\n");
}
return 0;
}
标签：printf \n return max sharp queue 转化 from algorithm
原文地址：https://www.cnblogs.com/DF-yimeng/p/9740863.html
