POJ2396 Budget
这题是真的恶心……
有源汇的上下界网络流求可行流……
首先矩阵的建图基本比较清晰，就是行列之间连边，其上下界由给定的条件决定。这题其实有两种改造法都能过。第一种是最正统的套路，就是首先建立原点和汇点，然后把行向原点连边，容量全都是0（因为上下界的差值是0），不过要更改这些点的流入和流出下限，中间的边还是按上界减去下界为容量。这样我们现在形成了一个有源汇的上下界网络流，只要加上一条从汇点到原点容量为INF的一条边，它就成了一个无源汇的上下界网络流。然后建立辅助源汇点跑流即可。
第二种比较简洁，就是不建立源汇点，只把矩阵行列之间的边构造出来之后，直接建立辅助源汇点跑流。这样其实也是正确的，为啥呢……因为其实第一种和第二种的不同在于，他多了源汇点……但是因为源汇点直接为了转换为无源汇的图，又加上了一条INF的边，这样的话流就可以无限的从这里跑了……所以第一种多出来的流最后也没有用上……（以上都是感性理解orz……其实我也不能详细证明为啥都行）
然后就这么建图这题就可以过了。但是这题的建图是真的恶心……尤其是如果你没有使用邻接矩阵存流的下届……我是用下标计算处理的……debug到痛不欲生……看一下两种代码吧。
第一种：
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')
#define fr friend inline
#define y1 poj
#define mp make_pair
#define pr pair<int,int>
#define fi first
#define sc second
#define pb push_back
using namespace std;
typedef long long ll;
const int M = 1000005;
const int N = 505;
const int INF = 0x3f3f3f3f;
const double eps = 1e-7;
int read()
{
int ans = 0,op = 1;char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') op = -1;ch = getchar();}
while(ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0',ch = getchar();
return ans * op;
}
struct edge
{
int next,to,from,v;
}e[M<<1];
int Ti,n,m,head[N<<1],ecnt,x,y,a[M<<1],b[M<<1];
int rsum[N<<1],csum[N<<1],S,T,tot,dep[N<<1],cur[N<<1];
int hsum[N<<1],lsum[N<<1],S1,T1,k,z;
queue <int> q;
char s[3];
bool flag;
void add(int x,int y,int z)
{
e[++ecnt].to = y;
e[ecnt].from = x;
e[ecnt].next = head[x];
e[ecnt].v = z;
head[x] = ecnt;
}
void build()
{
x = read(),y = read(),scanf("%s",s),z = read();
if(s[0] == '>')
{
z++;
if((z > hsum[x] && x != 0) || (z > lsum[y] && y != 0)) flag = 1;
if(x == 0 && y == 0) rep(i,1,n*m) a[i] = z;
else if(x == 0) rep(i,0,n-1) a[i*m+y] = z;
else if(y == 0) rep(i,1,m) a[(x-1)*m+i] = z;
else a[(x-1)*m+y] = z;
}
if(s[0] == '<')
{
z--;
if(x == 0 && y == 0) rep(i,1,n*m) b[i] = z;
else if(x == 0) rep(i,0,n-1) b[i*m+y] = z;
else if(y == 0) rep(i,1,m) b[(x-1)*m+i] = z;
else b[(x-1)*m+y] = z;
}
if(s[0] == '=')
{
if((z > hsum[x] && x != 0) || (z > lsum[y] && y != 0)) flag = 1;
if(x == 0 && y == 0) rep(i,1,n*m) a[i] = b[i] = z;
else if(x == 0) rep(i,0,n-1) a[i*m+y] = b[i*m+y] = z;
else if(y == 0) rep(i,1,m) a[(x-1)*m+i] = b[(x-1)*m+i] = z;
else a[(x-1)*m+y] = b[(x-1)*m+y] = z;
}
}
void rebuild()
{
rep(i,S,T)
{
int now = csum[i] - rsum[i];
if(now < 0) add(S1,i,-now),add(i,S1,0);
else add(i,T1,now),add(T1,i,0),tot += now;
}
}
void clear()
{
flag = 0;
memset(head,-1,sizeof(head)),ecnt = -1,tot = 0;
memset(a,0,sizeof(a)),memset(b,0x3f,sizeof(b));
rep(i,S,T1) rsum[i] = csum[i] = 0;
}
bool bfs(int s,int t)
{
while(!q.empty()) q.pop();
rep(i,0,t) cur[i] = head[i];
memset(dep,-1,sizeof(dep));
dep[s] = 0,q.push(s);
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = head[k];~i;i = e[i].next)
{
if(e[i].v && dep[e[i].to] == -1)
dep[e[i].to] = dep[k] + 1,q.push(e[i].to);
}
}
return dep[t] != -1;
}
int dfs(int s,int t,int lim)
{
if(s == t || !lim) return lim;
int flow = 0;
for(int i = cur[s];~i;i = e[i].next)
{
cur[s] = i;
if(dep[e[i].to] != dep[s] + 1) continue;
int f = dfs(e[i].to,t,min(lim,e[i].v));
if(f)
{
e[i].v -= f,e[i^1].v += f;
flow += f,lim -= f;
if(!lim) break;
}
}
if(!flow) dep[s] = -1;
return flow;
}
int dinic(int s,int t)
{
int maxflow = 0;
while(bfs(s,t)) maxflow += dfs(s,t,INF);
return maxflow;
}
int main()
{
//freopen("a.in","r",stdin);
//freopen("f.out","w",stdout);
Ti = read();
while(Ti--)
{
n = read(),m = read(),T = n + m + 1,S1 = T + 1,T1 = S1 + 1;
clear();
rep(i,1,n) hsum[i] = read();
rep(i,1,m) lsum[i] = read();
k = read();
rep(i,1,k) build();
rep(i,1,n)
rep(j,1,m)
{
int f = a[(i-1)*m+j],g = b[(i-1)*m+j];
add(i,j+n,g-f),add(j+n,i,0);
csum[i] += f,rsum[j+n] += f;
}
rep(i,1,n) add(S,i,0),add(i,S,0),csum[S] += hsum[i],rsum[i] += hsum[i];
rep(i,1,m) add(i+n,T,0),add(T,i+n,0),csum[i+n] += lsum[i],rsum[T] += lsum[i];
add(T,S,INF),add(S,T,0);
rebuild();
int g = dinic(S1,T1);
if(flag || tot > g) printf("IMPOSSIBLE\n");
else
{
rep(i,1,n)
{
rep(j,1,m)
{
int g = (i - 1) * m + j - 1;
printf("%d ",e[g<<1|1].v + a[g+1]);
}
enter;
}
}
}
return 0;
}
第二种：
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')
#define fr friend inline
#define y1 poj
#define mp make_pair
#define pr pair<int,int>
#define fi first
#define sc second
#define pb push_back
using namespace std;
typedef long long ll;
const int M = 1000005;
const int N = 505;
const int INF = 0x3f3f3f3f;
const double eps = 1e-7;
int read()
{
int ans = 0,op = 1;char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') op = -1;ch = getchar();}
while(ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0',ch = getchar();
return ans * op;
}
struct edge
{
int next,to,from,v;
}e[M<<1];
int Ti,n,m,head[N<<1],ecnt,x,y,a[M<<1],b[M<<1];
int rsum[N<<1],csum[N<<1],S,T,tot,dep[N<<1],cur[N<<1];
int hsum[N<<1],lsum[N<<1],S1,T1,k,z;
queue <int> q;
char s[3];
bool flag;
void add(int x,int y,int z)
{
e[++ecnt].to = y;
e[ecnt].from = x;
e[ecnt].next = head[x];
e[ecnt].v = z;
head[x] = ecnt;
}
void build()
{
x = read(),y = read(),scanf("%s",s),z = read();
if(s[0] == '>')
{
z++;
if((z > hsum[x] && x != 0) || (z > lsum[y] && y != 0)) flag = 1;
if(x == 0 && y == 0) rep(i,1,n*m) a[i] = z;
else if(x == 0) rep(i,0,n-1) a[i*m+y] = z;
else if(y == 0) rep(i,1,m) a[(x-1)*m+i] = z;
else a[(x-1)*m+y] = z;
}
if(s[0] == '<')
{
z--;
if(x == 0 && y == 0) rep(i,1,n*m) b[i] = z;
else if(x == 0) rep(i,0,n-1) b[i*m+y] = z;
else if(y == 0) rep(i,1,m) b[(x-1)*m+i] = z;
else b[(x-1)*m+y] = z;
}
if(s[0] == '=')
{
if((z > hsum[x] && x != 0) || (z > lsum[y] && y != 0)) flag = 1;
if(x == 0 && y == 0) rep(i,1,n*m) a[i] = b[i] = z;
else if(x == 0) rep(i,0,n-1) a[i*m+y] = b[i*m+y] = z;
else if(y == 0) rep(i,1,m) a[(x-1)*m+i] = b[(x-1)*m+i] = z;
else a[(x-1)*m+y] = b[(x-1)*m+y] = z;
}
}
void rebuild()
{
rep(i,S,T)
{
int now = csum[i] - rsum[i];
//printf("%d %d\n",i,now);
if(now < 0) add(S1,i,-now),add(i,S1,0);
else add(i,T1,now),add(T1,i,0),tot += now;
}
}
void clear()
{
flag = 0;
memset(head,-1,sizeof(head)),ecnt = -1,tot = 0;
memset(a,0,sizeof(a)),memset(b,0x3f,sizeof(b));
rep(i,S,T1) rsum[i] = csum[i] = 0;
}
bool bfs(int s,int t)
{
while(!q.empty()) q.pop();
rep(i,0,t) cur[i] = head[i];
memset(dep,-1,sizeof(dep));
dep[s] = 0,q.push(s);
while(!q.empty())
{
int k = q.front();q.pop();
for(int i = head[k];~i;i = e[i].next)
{
if(e[i].v && dep[e[i].to] == -1)
dep[e[i].to] = dep[k] + 1,q.push(e[i].to);
}
}
return dep[t] != -1;
}
int dfs(int s,int t,int lim)
{
if(s == t || !lim) return lim;
int flow = 0;
for(int i = cur[s];~i;i = e[i].next)
{
cur[s] = i;
if(dep[e[i].to] != dep[s] + 1) continue;
int f = dfs(e[i].to,t,min(lim,e[i].v));
if(f)
{
e[i].v -= f,e[i^1].v += f;
flow += f,lim -= f;
if(!lim) break;
}
}
if(!flow) dep[s] = -1;
return flow;
}
int dinic(int s,int t)
{
int maxflow = 0;
while(bfs(s,t)) maxflow += dfs(s,t,INF);
return maxflow;
}
int main()
{
//freopen("a.in","r",stdin);
//freopen("f.out","w",stdout);
Ti = read();
while(Ti--)
{
n = read(),m = read(),T = n + m + 1,S1 = T + 1,T1 = S1 + 1;
clear();
rep(i,1,n) hsum[i] = read(),rsum[i] += hsum[i];
rep(i,1,m) lsum[i] = read(),csum[i+n] += lsum[i];
k = read();
rep(i,1,k) build();
rep(i,1,n)
rep(j,1,m)
{
int f = a[(i-1)*m+j],g = b[(i-1)*m+j];
add(i,j+n,g-f),add(j+n,i,0);
csum[i] += f,rsum[j+n] += f;
}
rebuild();
int g = dinic(S1,T1);
if(flag || tot > g) printf("IMPOSSIBLE\n");
else
{
rep(i,1,n)
{
rep(j,1,m)
{
int g = (i - 1) * m + j - 1;
printf("%d ",e[g<<1|1].v + a[g+1]);
}
enter;
}
}
}
return 0;
}
当你意识到，每个上一秒都成为永恒。
