POJ 1273(HDU 1532) Drainage Ditches|HDU 3549 Flow Problem|很典型&&裸的网络流-优快云博客

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Drainage Ditches

**
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 64219 Accepted: 24777

Description

Every time it rains on Farmer John’s fields, a pond forms over Bessie’s favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie’s clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

5 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output

50
**

Source

USACO 93

简单题意

单向边的最大流 1 是起点 m是终点

Solution

**
起初Wa==》发现多组数据…

然后TLE 发现maxn定义等于200…

估计是被NOIP自己的成绩吓哭了…

裸题不多说…代码觉得还是不是很长的..QAQ

用的邻接表储存的如果希望看其他版本的代码点这里这里写链接内容

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn=205;

struct data{int to,next,w;}e[maxn*maxn];

int head[maxn],cnt=1,h[maxn],n,m,ans;

void ins(int u,int v,int w){cnt++;e[cnt].to=v;e[cnt].w=w;e[cnt].next=head[u];head[u]=cnt;}

bool bfs(){
    memset(h,-1,sizeof(h));
    h[1]=0;
    queue<int> q;
    q.push(1);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=head[x];i;i=e[i].next)
            if(e[i].w&&h[e[i].to]<0)
            {
                q.push(e[i].to);
                h[e[i].to]=h[x]+1;
            }
    }
    if(h[n]==-1)return 0;
    return 1;
}
int dfs(int x,int f){
    if(x==n)return f;
    int w,used=0;
    for(int i=head[x];i;i=e[i].next)
    if(e[i].w&&h[e[i].to]==h[x]+1)
    {
        w=f-used;
        w=dfs(e[i].to,min(w,e[i].w));
        e[i].w-=w;
        e[i^1].w+=w;
        used+=w;
        if(used==f)return f;
    }
    if(!used)h[x]=-1;
    return used;
}
void Dinic(){
    while(bfs()){

        ans+=dfs(1,2000000000);
    }
}
int main()
{

    while(scanf("%d%d",&m,&n)==2)
    {
        ans=0;cnt=1;
        memset(head,0,sizeof(head));
    for(int i=1;i<=m;i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        ins(u,v,w);
        ins(v,u,0);
    }
    Dinic();
    printf("%d\n",ans); 
    }
    return 0;
}

HDU 3549

这是后来加的…… 一样……就改了改细节……

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int maxn=205;

struct data{int to,next,w;}e[maxn*maxn];

int head[maxn],cnt=1,h[maxn],n,m,ans,T;

void ins(int u,int v,int w){cnt++;e[cnt].to=v;e[cnt].w=w;e[cnt].next=head[u];head[u]=cnt;}

bool bfs(){
    memset(h,-1,sizeof(h));
    h[1]=0;
    queue<int> q;
    q.push(1);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=head[x];i;i=e[i].next)
            if(e[i].w&&h[e[i].to]<0)
            {
                q.push(e[i].to);
                h[e[i].to]=h[x]+1;
            }
    }
    if(h[n]==-1)return 0;
    return 1;
}
int dfs(int x,int f){
    if(x==n)return f;
    int w,used=0;
    for(int i=head[x];i;i=e[i].next)
    if(e[i].w&&h[e[i].to]==h[x]+1)
    {
        w=f-used;
        w=dfs(e[i].to,min(w,e[i].w));
        e[i].w-=w;
        e[i^1].w+=w;
        used+=w;
        if(used==f)return f;
    }
    if(!used)h[x]=-1;
    return used;
}
void Dinic(){
    while(bfs()){

        ans+=dfs(1,2000000000);
    }
}
int main()
{
    int T,Cnt=0;
    scanf("%d",&T);
    while(T--&&scanf("%d%d",&n,&m)==2)
    {
        Cnt++;
        ans=0;cnt=1;
        memset(head,0,sizeof(head));
    for(int i=1;i<=m;i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        ins(u,v,w);
        ins(v,u,0);
    }
    Dinic();
    printf("Case %d: %d\n",Cnt,ans);
    }
    return 0;
}