POJ1149 PIGS 网络最大流

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can't unlock any pighouse because he doesn't have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of
pigs.

All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.

More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across
the unlocked pig-houses.

An unlimited number of pigs can be placed in every pig-house.

Write a program that will find the maximum number of pigs that he can sell on that day.
Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.

The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.

The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):

A K1 K2 ... KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, ..., KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.
Output

The first and only line of the output should contain the number of sold pigs.
Sample Input

3 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

一般增广路径方法——Ford-Fulkerson算法（标号法）

468K

0MS

G++

1601B

#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f

int s, t, m, n;
int c[105][105], f[105][105];
int house[1005], pre[1005];
int queue[105], qs, qe;
int mark[105], minf[105];

void init(){
int num, k;
memset(pre, 0, sizeof(pre));
memset(c, 0, sizeof(c));
scanf("%d %d", &m, &n);
s = 0; t = n + 1;
for(int i = 1; i <= m; i++)
scanf("%d", house + i);
for(int i = 1; i <= n; i++){
scanf("%d", &num);
while(num--){
scanf("%d", &k);
if(!pre[k])
c[s][i] += house[k];
else
c[ pre[k] ][i] = INF;
pre[k] = i;
}
scanf("%d", &c[i][t]);
}
}

void ford(){
int v, p;
memset(f, 0, sizeof(f));
minf[0] = INF;

while(true){
for(int i = 0; i <= n + 1; i++)
mark[i] = -2;
mark[0] = -1;

qs = 0; queue[qs] = 0; qe = 1;
while( qs != qe && mark[t] == -2){
v = queue[qs];  qs = (qs + 1) % n;
for(int i = 0; i <= t; i ++)
if( mark[i] == -2 && (p = c[v][i] - f[v][i]) ){
mark[i] = v; queue[qe] = i; qe = (qe + 1) % n;
minf[i] = minf[v] < p ? minf[v] : p;
}
}
if(mark[t] == -2) break;

for(int i = mark[t], j = t;  i != -1;  j = i,i = mark[i] ){
f[i][j] += minf[t];
f[j][i] -= f[i][j];
}
}

int ans = 0;
for(int i = 1; i <= n; i++)
ans += f[i][t];
printf("%d\n", ans);
}

int main(){
init();
ford();
return 0;
}

最短增广路算法——SAP

396K

16MS

G++

1673B

#include <cstdio>
#include <cstring>
#define inf 0x3f3f3f3f

int n, m, s, t;
int c[105][105], h[105], vh[105];

void init(){
scanf("%d %d", &m, &n);
int a[m], b[m];
memset(a, 0, sizeof(a));
memset(vh, 0, sizeof(vh));
memset(h, 0, sizeof(h));
memset(c, 0, sizeof(c));
for(int i = 0; i < m; i++)
scanf("%d", &b[i]);
for(int i = 0; i < n; i++){
int k, l, temp;
scanf("%d", &k);
for(int j = 0; j < k; j++){
scanf("%d", &temp);
if(!a[temp - 1])
c[0][i+1] += b[temp-1];
else
c[ a[temp-1] ][i+1] = inf;
a[temp-1] = i + 1;
}
scanf("%d", &l);
c[i+1][n+1] = l;
}
}

int sap(int i, int f){
if(i == t)
return f;
int old = f, minh = n - 1;
for(int j = 0; j < n; j++){
if(c[i][j] > 0){
if(h[i] == h[j] + 1){
int d = c[i][j] < old ? c[i][j] : old;
d = sap(j, d);
c[i][j] -= d;
c[j][i] += d;
old -= d;
if(h[s] >= n)
return  f - old;
if(old == 0)
break;
}
if(h[j] < minh)
minh = h[j];
}
}
if(f == old){
vh[ h[i] ]--;
if(vh[ h[i] ] == 0)
h[s] = n;
h[i] = minh+1;
vh[ h[i] ] ++;
}
return f - old;
}

int main(){
init();
s = 0, t = n + 1, vh[0] = n + 2;
int ff = 0;
n += 2;
while(h[s] < n)
ff += sap(s, inf);
printf("%d\n", ff);
return 0;
}

连续最短增广路算法——Dinic

400K

0MS

G++

2270B

#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f

int n, m, s, t;
int house[1005], pre[1005];
int c[105][105], ps[105], dep[105];

void init(){
int num, k;
memset(pre, 0, sizeof(pre));
memset(c, 0, sizeof(c));
scanf("%d %d", &m, &n);
s = 0; t = n + 1;
for(int i = 1; i <= m; i++)
scanf("%d", house + i);
for(int i = 1; i <= n; i++){
scanf("%d", &num);
while(num--){
scanf("%d", &k);
if(!pre[k])
c[s][i] += house[k];
else
c[ pre[k] ][i] = INF;
pre[k] = i;
}
scanf("%d", &c[i][t]);
}
}

bool bfs(int n,int s,int t){
int f = 0, r = 0, u, v;
memset(dep, -1, sizeof(dep));
ps[r++] = s;
dep[s] = 0; //开始计算层次
while(f < r){
u = ps[f++];
for(v = 0; v < n; v++)
if(c[u][v] && dep[v] < 0){
dep[v] = dep[u] + 1;
ps[r++] = v;
}
if(u == t) break;

}
return dep[t] < 0; //返回能否增广
}

int Dinic(int n,int s,int t){
int k, u, v, num, res = 0, top, tag;
while(1){
if( bfs(n, s, t) ) break; //不能增广，最终结果
top = 0;
u = s;
while(1){
if(u == t){  //一条增广
ps[top++] = t;
for(k = 0, num = INF; k < top - 1; k++)
if(c[ ps[k] ][ ps[k+1] ] < num){
tag = k; //标志最小弧的头
num = c[ ps[k] ][ ps[k+1] ];
}
for(k=0;k<top-1;k++){
c[ps[k]][ps[k+1]] -= num;
c[ps[k+1]][ps[k]] += num;
}
res += num;
top = tag; //从标志处继续找其他增广路
u = tag;
}
for(v = 0; v < n; v++)
if(c[u][v] && dep[u] + 1 == dep[v]) break;  //从u出发存在弧u->v

if(v<n) {ps[top++] = u; u = v;}   //压u入栈，从v出发找
else{
if(top == 0) break;  //确定增广路找不到
dep[u] = -1;
u=ps[--top]; //把u弹出，找层次图中另一条u->v
}
}
}
return res;
}

int main(){
init();
printf("%d\n", Dinic(n + 2, s, t));
return 0;
}