| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 5412 | Accepted: 1863 | Special Judge | ||
Description
As you know, all the computers used for ACM contests must be identical, so the participants compete on equal terms. That is why all these computers are historically produced at the same factory.
Every ACM computer consists of P parts. When all these parts are present, the computer is ready and can be shipped to one of the numerous ACM contests.
Computer manufacturing is fully automated by using N various machines. Each machine removes some parts from a half-finished computer and adds some new parts (removing of parts is sometimes necessary as the parts cannot be added to a computer in arbitrary order). Each machine is described by its performance (measured in computers per hour), input and output specification.
Input specification describes which parts must be present in a half-finished computer for the machine to be able to operate on it. The specification is a set of P numbers 0, 1 or 2 (one number for each part), where 0 means that corresponding part must not be present, 1 — the part is required, 2 — presence of the part doesn't matter.
Output specification describes the result of the operation, and is a set of P numbers 0 or 1, where 0 means that the part is absent, 1 — the part is present.
The machines are connected by very fast production lines so that delivery time is negligibly small compared to production time.
After many years of operation the overall performance of the ACM Computer Factory became insufficient for satisfying the growing contest needs. That is why ACM directorate decided to upgrade the factory.
As different machines were installed in different time periods, they were often not optimally connected to the existing factory machines. It was noted that the easiest way to upgrade the factory is to rearrange production lines. ACM directorate decided to entrust you with solving this problem.
Input
Input file contains integers P N, then N descriptions of the machines. The description of ith machine is represented as by 2 P + 1 integers Qi Si,1 Si,2...Si,P Di,1 Di,2...Di,P, where Qi specifies performance,Si,j — input specification for part j, Di,k — output specification for part k.
Constraints
1 ≤ P ≤ 10, 1 ≤ N ≤ 50, 1 ≤ Qi ≤ 10000
Output
Output the maximum possible overall performance, then M — number of connections that must be made, then M descriptions of the connections. Each connection between machines A and B must be described by three positive numbers A B W, where W is the number of computers delivered from A to B per hour.
If several solutions exist, output any of them.
Sample Input
Sample input 1 3 4 15 0 0 0 0 1 0 10 0 0 0 0 1 1 30 0 1 2 1 1 1 3 0 2 1 1 1 1 Sample input 2 3 5 5 0 0 0 0 1 0 100 0 1 0 1 0 1 3 0 1 0 1 1 0 1 1 0 1 1 1 0 300 1 1 2 1 1 1 Sample input 3 2 2 100 0 0 1 0 200 0 1 1 1
Sample Output
Sample output 1 25 2 1 3 15 2 3 10 Sample output 2 4 5 1 3 3 3 5 3 1 2 1 2 4 1 4 5 1 Sample output 3 0 0
Hint
Source
题意：一个电脑由n个部件组成，如今有m台机器，每台机器能够将一个组装状态的电脑组合成还有一个状态。如(0, 1, 2)表示第一个部件未完毕。第二个部件完毕，第三个部件可完毕可不完毕。然后给出m个机器单位时间内能完毕的任务数以及详细的输入和输出状态。求整个系统单位时间内的电脑成品产量以及详细的机器间的传输关联。
题解：这题能够转换成最大流来做，一个机器的输出状态能够跟还有一个机器的输入状态关联，仅仅要它们的状态“equals”，然后再设置一个超级源点和汇点，再就能够用Dinic解题了。
#include <stdio.h>
#include <string.h>
#define maxn 55
#define inf 0x3fffffff
struct Node {
int in[10], out[10]; // 拆点
int Q; // 容量
} M[maxn];
int G[maxn << 1][maxn << 1], que[maxn << 1], m, n, mp;
int G0[maxn << 1][maxn << 1], deep[maxn << 1], vis[maxn << 1];
bool equals(int a[], int b[]) {
for(int k = 0; k < n; ++k) {
if(a[k] != 2 && b[k] != 2 && a[k] != b[k])
return false;
}
return true;
}
bool countLayer() {
int i, id = 0, now, front = 0;
memset(deep, 0, sizeof(deep));
deep[0] = 1; que[id++] = 0;
while(front < id) {
now = que[front++];
for(i = 0; i <= mp; ++i)
if(G[now][i] && !deep[i]) {
deep[i] = deep[now] + 1;
if(i == mp) return true;
que[id++] = i;
}
}
return false;
}
int Dinic() {
int i, id = 0, maxFlow = 0, minCut, pos, u, v, now;
while(countLayer()) {
memset(vis, 0, sizeof(vis));
vis[0] = 1; que[id++] = 0;
while(id) {
now = que[id - 1];
if(now == mp) {
minCut = inf;
for(i = 1; i < id; ++i) {
u = que[i - 1]; v = que[i];
if(G[u][v] < minCut) {
minCut = G[u][v]; pos = u;
}
}
maxFlow += minCut;
for(i = 1; i < id; ++i) {
u = que[i - 1]; v = que[i];
G[u][v] -= minCut;
G[v][u] += minCut;
}
while(id && que[id - 1] != pos)
vis[que[--id]] = 0;
} else {
for(i = 0; i <= mp; ++i) {
if(G[now][i] && deep[now] + 1 == deep[i] && !vis[i]) {
que[id++] = i; vis[i] = 1; break;
}
}
if(i > mp) --id;
}
}
}
return maxFlow;
}
int main() {
//freopen("stdin.txt", "r", stdin);
int i, j, sum, count;
while(scanf("%d%d", &n, &m) == 2) {
memset(G, 0, sizeof(G));
for(i = 1; i <= m; ++i) {
scanf("%d", &M[i].Q);
for(j = 0; j < n; ++j) scanf("%d", &M[i].in[j]);
for(j = 0; j < n; ++j) scanf("%d", &M[i].out[j]);
G[i][i + m] = M[i].Q;
}
// 连接出口跟入口
for(i = 1; i <= m; ++i) {
for(j = i + 1; j <= m; ++j) {
if(equals(M[i].out, M[j].in))
G[i + m][j] = inf;
if(equals(M[j].out, M[i].in))
G[j + m][i] = inf;
}
}
// 设置超级源点和超级汇点
for(i = 1; i <= m; ++i) { // 源点
G[0][i] = inf;
for(j = 0; j < n; ++j)
if(M[i].in[j] == 1) {
G[0][i] = 0; break;
}
}
mp = m << 1 | 1;
for(i = 1; i <= m; ++i) { // 汇点
G[i + m][mp] = inf;
for(j = 0; j < n; ++j)
if(M[i].out[j] != 1) {
G[i + m][mp] = 0; break;
}
}
// 备份原图
memcpy(G0, G, sizeof(G));
sum = Dinic();
count = 0;
// 推断哪些路径有流走过
for(i = m + 1; i < mp; ++i)
for(j = 1; j <= m; ++j)
if(G0[i][j] > G[i][j]) ++count;
printf("%d %d
", sum, count);
// 输出机器间的关系
if(count)
for(i = m + 1; i < mp; ++i)
for(j = 1; j <= m; ++j)
if(G0[i][j] > G[i][j])
printf("%d %d %d
", i - m, j, G0[i][j] - G[i][j]);
}
return 0;
}2015.4.20
#include <stdio.h>
#include <string.h>
#include <vector>
using std::vector;
const int maxn = 102;
const int maxp = 10;
const int maxm = 2500;
const int inf = 0x3f3f3f3f;
const int sOut[maxp] = {};
int P, N;
struct Node2 {
int in[maxp], out[maxp];
int c;
} node[maxn];
int G[maxn][maxn], G0[maxn][maxn], queue[maxn];
bool vis[maxn]; int Layer[maxn];
bool countLayer(int s, int t) {
memset(Layer, 0, sizeof(Layer));
int id = 0, front = 0, now, i;
Layer[s] = 1; queue[id++] = s;
while(front < id) {
now = queue[front++];
for(i = s; i <= t; ++i)
if(G[now][i] && !Layer[i]) {
Layer[i] = Layer[now] + 1;
if(i == t) return true;
else queue[id++] = i;
}
}
return false;
}
// 源点，汇点，源点编号必须最小。汇点编号必须最大
int Dinic(int s, int t) {
int minCut, pos, maxFlow = 0;
int i, id = 0, u, v, now;
while(countLayer(s, t)) {
memset(vis, 0, sizeof(vis));
vis[s] = true; queue[id++] = s;
while(id) {
now = queue[id - 1];
if(now == t) {
minCut = inf;
for(i = 1; i < id; ++i) {
u = queue[i - 1];
v = queue[i];
if(G[u][v] < minCut) {
minCut = G[u][v];
pos = u;
}
}
maxFlow += minCut;
for(i = 1; i < id; ++i) {
u = queue[i - 1];
v = queue[i];
G[u][v] -= minCut;
G[v][u] += minCut;
}
while(queue[id - 1] != pos)
vis[queue[--id]] = false;
} else {
for(i = 0; i <= t; ++i) {
if(G[now][i] && Layer[now] + 1 == Layer[i] && !vis[i]) {
vis[i] = 1; queue[id++] = i; break;
}
}
if(i > t) --id;
}
}
}
return maxFlow;
}
void init()
{
memset(G, 0, sizeof(G));
}
bool canBeLinked(const int out[], const int in[])
{
int i;
for (i = 0; i < P; ++i) {
if (out[i] == 0 && in[i] == 1) break;
if (out[i] == 1 && in[i] == 0) break;
}
return i == P;
}
void getMap()
{
int u, v, c, i, j, k, cnt;
for (i = 1; i <= N; ++i) {
scanf("%d", &c);
G[i][N+i] = c;
for (j = 0; j < P; ++j)
scanf("%d", &node[i].in[j]);
for (j = cnt = 0; j < P; ++j) {
scanf("%d", &node[i].out[j]);
if (node[i].out[j] == 1) ++cnt;
}
if (canBeLinked(sOut, node[i].in)) G[0][i] = inf;
if (cnt == P) G[i+N][2*N+1] = inf;
}
for (i = 1; i <= N; ++i) {
// connection
for (j = 1; j <= N; ++j) {
if (i != j && canBeLinked(node[i].out, node[j].in)) {
G[i+N][j] = inf;
// printf("...%d..%d...
", i + N, j);
}
}
}
memcpy(G0, G, sizeof(G));
/*
for (i = 1; i <= N; ++i) {
printf("...%d: ", i);
for (j = 0; j < P; ++j)
printf("%d%c", node[i].in[j], j == P - 1 ? '
' : ' ');
for (j = 0; j < P; ++j)
printf("%d%c", node[i].out[j], j == P - 1 ? '
' : ' ');
}
for (i = 0; i <= 2 * N + 1; ++i) {
printf("%d: ", i);
for (j = 0; j <= 2 * N + 1; ++j) {
printf("%d%c", G[i][j], j == 2*N+1 ? '
' : ' ');
}
}*/
}
int solve()
{
int ret = Dinic(0, 2 * N + 1);
vector<int> vec;
int cnt = 0, i, j, u, v, c;
for (i = 1; i <= N; ++i) {
for (j = 1; j <= N; ++j) {
if (G[i+N][j] < G0[i+N][j]) {
vec.push_back(i);
vec.push_back(j);
vec.push_back(G0[i+N][j] - G[i+N][j]);
}
}
}
printf("%d %d
", ret, vec.size() / 3);
for (i = 0; i < vec.size(); ) {
u = vec[i++];
v = vec[i++];
c = vec[i++];
printf("%d %d %d
", u, v, c);
}
}
int main()
{
while (~scanf("%d%d", &P, &N)) {
init();
getMap();
solve();
}
return 0;
}