Problem Portal
Portal1: Luogu
Description
If that, given a network diagram, and the source and sink, the network obtains its maximum flow.
Input
The first row contains four integer \ (N, M, S, T \) , respectively, the number of points, there are the number of edges, number source, sink point number.
Next \ (M \) lines contains three positive integers \ (u_i, V_I, W_i \) , represents the \ (I \) Article directed edge from \ (W_i \) arriving \ (V_I \) , the right side is \ (w_i \) (that is, the maximum flow of side \ (w_i \) ).
Output
Line, contains a positive integer, is the maximum flow of the network.
Sample Input
4 5 4 3
4 2 30
4 3 20
2 3 20
2 1 30
1 3 40
Sample Output
50
Hint
Data Scale:
For \ (30 \% \) Data: \ (N \ Leq 10, M \ Leq 25 \) ;
For \ (70 \% \) Data: \ (N \ Leq 200 is, M \ Leq 1000 \) ;
For \ (100 \% \) Data: \ (N \ Leq 10000, M \ Leq 100000 \) .
Sample Description:
Present title \ (3 \) paths:
\ (. 4 \ to 2 \ to. 3 \) , the route through \ (20 \) flow
\ (. 4 \. 3 to \) , by \ (20 \) flow
\ (. 4 \ to 2 \ to. 1 \. 3 to \) , by \ (10 \) flow (side \ (4 \ 2 \) to have previously been spent \ (20 \) flow rate)
Therefore, the total flow rate \ (20 + 20 + 10 = 50 \) , output \ (50 \) .
Solution
Template questions, seeking maximum flow.
Code
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f, MAXN=10005, MAXM=200005;
int n, m, s, t, u, v, val, cnt, head[MAXN], dis[MAXN], pre[MAXN];
queue<int> Q;
struct node {
int u, v, val, flow, nxt;
} edge[MAXM];
inline void addedge(int u, int v, int w) {//前向星存图
edge[++cnt].u=u; edge[cnt].v=v; edge[cnt].val=w; edge[cnt].nxt=head[u]; head[u]=cnt;
}
inline bool Emonds_Karp() {
memset(pre, 0, sizeof(pre));
memset(dis, 0, sizeof(dis));//初始化
dis[s]=INF;
pre[s]=0;
Q.push(s);
while(!Q.empty()) {//类似于SPFA
int x=Q.front();
Q.pop();
for (int i=head[x]; i; i=edge[i].nxt) {
int v=edge[i].v;
if (!dis[v] && edge[i].val>edge[i].flow) {
Q.push(v);
dis[v]=min(dis[x], edge[i].val-edge[i].flow);
pre[v]=i;
}
if (dis[t]) break;
}
}
return dis[t];
}
int main() {
scanf("%d%d%d%d",&n, &m, &s, &t);
cnt=1;
for(int i=1; i<=m; i++) {
scanf("%d%d%d",&u, &v, &val);
addedge(u, v, val);
addedge(v, u, 0);//前向星存图
}
int ans=0;
while (Emonds_Karp()) {//Emonds Karp算法
for (int i=t; i!=s; i=edge[pre[i]].u) {
edge[pre[i]].flow+=dis[t];
edge[pre[i] xor 1].val-=dis[t];
}
ans+=dis[t];
}
printf("%d\n",ans);
return 0;
}
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f, MAXN=10005, MAXM=200005;
struct node {
int nxt, to, val;
} edge[MAXM];
queue<int> Q;
int n, m, s, t, u, v, cnt, val, dis[MAXN], head[MAXN];
inline void addedge(int u, int v, int w) {//前向星存图
edge[++cnt].val=w; edge[cnt].to=v; edge[cnt].nxt=head[u]; head[u]=cnt;
}
inline bool bfs() {
memset(dis, -1, sizeof(dis));
dis[s]=0;
Q.push(s);
while (!Q.empty()) {
int x=Q.front();
Q.pop();
for (int i=head[x]; ~i; i=edge[i].nxt) {
int v=edge[i].to;
if (edge[i].val && dis[v]==-1) {
dis[v]=dis[x]+1;
Q.push(v);
}
}
}
if (~dis[t]) return 1;
return 0;
}
inline int dfs(int u, int flow) {
if (u==t) return flow;
int ret=flow;
for (int i=head[u]; ~i; i=edge[i].nxt) {
if (ret<=0) break;
int v=edge[i].to;
if (edge[i].val && dis[v]==dis[u]+1) {
int x=dfs(v, min(edge[i].val, ret));
ret-=x;
edge[i].val-=x;
edge[i xor 1].val+=x;
}
}
return flow-ret;
}
inline int Dinic() {//Dinic算法
int ret=0;
while (bfs()) ret+=dfs(s, INF);
return ret;
}
int main() {
scanf("%d%d%d%d",&n, &m, &s, &t);
memset(head, -1, sizeof(head));
cnt=1;
for (int i=1; i<=m; i++) {
scanf("%d%d%d",&u, &v, &val);
addedge(u, v, val);
addedge(v, u, 0);//双边
}
printf("%d\n",Dinic());
return 0;
}
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f, MAXN=10005, MAXM=200005;
queue<int> Q;
int n, m, s, t, u, v, val, cnt, ans, head[MAXN], head1[MAXN], deep[MAXN], pre[MAXN], a[MAXN];
struct node {
int v, w, nxt;
} edge[MAXM];
inline void addedge(int u, int v, int w) {//前向星存图
edge[cnt].v=v; edge[cnt].w=w; edge[cnt].nxt=head[u]; head[u]=cnt++;
}
inline void bfs(int t) {
for (int i=1; i<=n; i++)
head1[i]=head[i];
for (int i=1; i<=n; i++)
deep[i]=n;
deep[t]=0;
Q.push(t);
while (!Q.empty()) {
int u=Q.front();
Q.pop();
for (int i=head[u]; ~i; i=edge[i].nxt)
if (deep[edge[i].v]==n && edge[i^1].w) {
deep[edge[i].v]=deep[u]+1;
Q.push(edge[i].v);
}
}
}
int calc(int s,int t) {//计算
int ans=INF, u=t;
while (u!=s) {
ans=min (ans,edge[pre[u]].w);
u=edge[pre[u] xor 1].v;
}
u=t;
while (u!=s) {
edge[pre[u]].w-=ans;
edge[pre[u] xor 1].w+=ans;
u=edge[pre[u] xor 1].v;
}
return ans;
}
inline void ISAP(int s, int t) {//ISAP算法
int u=s;
bfs(t);
for (int i=1; i<=n; i++)
a[deep[i]]++;
while (deep[s]<n) {
if (u==t) {
ans+=calc(s, t);
u=s;
}
bool flag=0;
for (int &i=head1[u]; ~i; i=edge[i].nxt)
if (deep[u]==deep[edge[i].v]+1 && edge[i].w) {
flag=1;
u=edge[i].v;
pre[edge[i].v]=i;
break;
}
if (!flag) {
int Min=n-1;
for (int i=head[u]; ~i; i=edge[i].nxt)
if (edge[i].w) Min=min(Min, deep[edge[i].v]);
if ((--a[deep[u]])==0) break;
a[deep[u]=Min+1]++;
head1[u]=head[u];
if (u!=s) u=edge[pre[u] xor 1].v;
}
}
}
int main() {
memset(head, -1, sizeof (head));
scanf("%d%d%d%d",&n, &m, &s, &t);
for (int i=1; i<=m; i++) {
scanf("%d%d%d",&u, &v, &val);
addedge(u, v, val);
addedge(v, u, 0);//双边
}
ISAP(s, t);
printf ("%d\n",ans);
return 0;
}