POJ 1698 Alice's Chance（最大流+拆點），poj1698

POJ 1698 Alice's Chance（最大流+拆點），poj1698
POJ 1698 Alice's Chance

題目連結

題意：拍n部電影，每部電影要在前w星期完成，並且一周只有一些天是可以拍的，每部電影有個需要的總時間，問是否能拍完電影

思路：源點向每部電影連邊，容量為d，然後每部電影對應能拍的那天連邊，由於每天容量限制是1，所以進行拆點，然後連向匯點即可

代碼：

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXNODE = 1005;const int MAXEDGE = 100005;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge {int u, v;Type cap, flow;Edge() {}Edge(int u, int v, Type cap, Type flow) {this->u = u;this->v = v;this->cap = cap;this->flow = flow;}};struct Dinic {int n, m, s, t;Edge edges[MAXEDGE];int first[MAXNODE];int next[MAXEDGE];bool vis[MAXNODE];Type d[MAXNODE];int cur[MAXNODE];vector<int> cut;void init(int n) {this->n = n;memset(first, -1, sizeof(first));m = 0;}void add_Edge(int u, int v, Type cap) {edges[m] = Edge(u, v, cap, 0);next[m] = first[u];first[u] = m++;edges[m] = Edge(v, u, 0, 0);next[m] = first[v];first[v] = m++;}bool bfs() {memset(vis, false, sizeof(vis));queue<int> Q;Q.push(s);d[s] = 0;vis[s] = true;while (!Q.empty()) {int u = Q.front(); Q.pop();for (int i = first[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (!vis[e.v] && e.cap > e.flow) {vis[e.v] = true;d[e.v] = d[u] + 1;Q.push(e.v);}}}return vis[t];}Type dfs(int u, Type a) {if (u == t || a == 0) return a;Type flow = 0, f;for (int &i = cur[u]; i != -1; i = next[i]) {Edge& e = edges[i];if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {e.flow += f;edges[i^1].flow -= f;flow += f;a -= f;if (a == 0) break;}}return flow;}Type Maxflow(int s, int t) {this->s = s; this->t = t;Type flow = 0;while (bfs()) {for (int i = 0; i < n; i++)cur[i] = first[i];flow += dfs(s, INF);}return flow;}void MinCut() {cut.clear();for (int i = 0; i < m; i += 2) {if (vis[edges[i].u] && !vis[edges[i].v])cut.push_back(i);}}} gao;int t, n, day[10], d, w, vis[400];int main() {scanf("%d", &t);while (t--) {memset(vis, 0, sizeof(vis));gao.init(1000);scanf("%d", &n);int sum = 0;for (int i = 1; i <= n; i++) {for (int j = 1; j <= 7; j++)scanf("%d", &day[j]);scanf("%d%d", &d, &w);sum += d;gao.add_Edge(0, i, d);for (int j = 0; j < w; j++) {for (int k = 1; k <= 7; k++) {if (day[k]) {gao.add_Edge(i, 20 + j * 7 + k, INF);vis[20 + j * 7 + k] = 1;}}}}for (int i = 21; i <= 370; i++) {if (!vis[i]) continue;gao.add_Edge(i, i + 350, 1);gao.add_Edge(i + 350, 1000, INF);}printf("%s\n", gao.Maxflow(0, 1000) == sum ? "Yes" : "No");}return 0;}

誰會用鄰接矩陣做Poj3204（最大流加枚舉）（做對了會加高分）

剛剛試過了 鄰接矩陣會TLE
因為樸素網路流是3方的
稍微改一下成鄰接表就好了
而且網路流基本不用鄰接矩陣了 都是用sap模板 不然資料一上千絕對掛的
#include <iostream>
using namespace std;
const int maxn=510;
const int maxm=5010;
const int maxl=99999999;
int a[maxm],c[maxn];
int f[maxn][maxn],g[maxn][maxn];
int l[maxn][maxn];
int r[maxn];
int c1[maxn],c2[maxn];
int m,n;
int e;
void iin()
{
int i,j,k,s;
int c[maxm];
scanf("%d%d",&n,&m);
memset(f,0,sizeof(f));
memset(g,0,sizeof(g));
memset(c,0,sizeof(c));
memset(r,0,sizeof(r));
for (i=0;i<m;i++)
{
scanf("%d%d%d",&j,&k,&s);
l[j][r[j]++]=k;
l[k][r[k]++]=j;
f[j][k]+=s;
g[j][k]+=s;
}
e=1;
for (i=0;i<n;i++)
while (e<f[0][i])
e*=2;
}
int min(int i,int j)
{
if (i>j)return j;else return i;
}
int dfs(int k,int p,int* c,int final)
{
int i,j,t;
if (k==final) return p;
c[k]=1;
for (t=0;t<r[k];t++)
{
i=l[k][t];
if (!c[i]&&f[k][i]>=e)
if (j=dfs(i,min(p,f[k][i]),c,final))
{
f[k][i]-=j;
f[i][k]+=j;
return j;
}
}
return 0;
}
void maxflow()
{
int tot=0,i,j;
while (e>0)
{
memset(c,0,sizeof(c));
while (i=dfs(0,maxl,c,n-1))
{
tot+=i;
memset(c,0,sizeof(c));
}
e/=2;
}
e=1;
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
dfs(0,maxl,c1,n-1);
for (i=0;i<n;i++) for (j=i+1;j<n;j++) {int code=f[i][j];f[i][j]=f[j][i];f[j][i]=code;}
dfs(n-1,maxl,c2,0);
tot=0;
for (i=0;i<n;i++) for (j=i+1;j<n;j++) {int code=f[i][j];f[i][j]=f[j][i];f[j][i]=code;}
for (i=0;i<n;i++) for (j=0;j&l......餘下全文>>