P3376 模板网络最大流——————Q - Marriage Match IV（最短路&最大流）

Posted 2020-11-23 kongbursi-2292702937

第一道题是模板题，下面主要是两种模板，但都用的是Dinic算法（第二个题也是）

第一题：

题意就不需要讲了，直接上代码：

vector代码：

  1 //invalid types ‘int[int]‘ for array subscript :字母重复定义
  2 #include<stdio.h>
  3 #include<string.h>
  4 #include<iostream>
  5 #include<stdlib.h>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<vector>
  9 using namespace std;
 10 const int N=10050;
 11 const int M=100050;
 12 const int INF=0x3f3f3f3f;
 13 int n,m,a,b,dep[N];
 14 struct shudui
 15 {
 16     int start,value;
 17 }str1;
 18 vector<shudui>w[N];
 19 queue<int>r;
 20 bool bfs()  //对每个点进行分层
 21 {
 22     for(int i=1;i<=n;++i) dep[i]=0;
 23     while(!r.empty())
 24     {
 25         r.pop();
 26     }
 27     r.push(a);
 28     dep[a]=1; //从起点开始增广
 29     while(!r.empty())
 30     {
 31         int now=r.front();
 32         r.pop();
 33         int len=w[now].size();
 34         for(int i=0;i<len;++i)
 35         {
 36             str1=w[now][i];
 37             if(str1.value && !dep[str1.start])
 38             {
 39                 dep[str1.start]=dep[now]+1;
 40                 r.push(str1.start);
 41             }
 42         }
 43     }
 44     return dep[b];
 45 }
 46 int dfs(int now,int aim,int flow)
 47 {
 48     if(now==aim || (!flow)) return flow;
 49     int res=0;
 50     int len=w[now].size();
 51     for(int i=0;i<len;++i)
 52     {
 53         str1=w[now][i];
 54         if(str1.value && dep[str1.start]==dep[now]+1)
 55         {
 56             int kk=dfs(str1.start,aim,min(flow,str1.value));
 57             res+=kk;
 58             flow-=kk;
 59             w[now][i].value-=kk;  //我突然方法你把vector中的值赋给str1的时候，你对石头人
 60             //但是vector中的值仍然没有变
 61             int len1=w[str1.start].size();
 62             for(int j=0;j<len1;++j)
 63             {
 64                 if(w[str1.start][j].start==now)
 65                 {
 66                     w[str1.start][j].value+=kk;
 67                     break;
 68                 }
 69             }
 70         }
 71     }
 72     return res;
 73 }
 74 int Dinic()
 75 {
 76     int ans=0;
 77     while(bfs())
 78     {
 79         //printf("**
");
 80         ans+=dfs(a,b,INF);
 81     }
 82     return ans;
 83 }
 84 int main()
 85 {
 86     scanf("%d%d%d%d",&n,&m,&a,&b);
 87     while(m--)
 88     {
 89         int u,v,ww;
 90         scanf("%d%d%d",&u,&v,&ww);
 91         str1.start=v;
 92         str1.value=ww;
 93         w[u].push_back(str1);
 94         str1.start=u;
 95         str1.value=0;   //它的方向边初始化为0
 96         w[v].push_back(str1);
 97     }
 98     //printf("**
");
 99     printf("%d
",Dinic());
100     return 0;
101 }

View Code

 链接表：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<queue>
 4 #include<vector>
 5 #include<algorithm>
 6 #include<iostream>
 7 #define N 10050
 8 #define M 100050
 9 using namespace std;
10 const int INF=0x3f3f3f3f;
11 int n,m,a,b,fir[N],dep[N],cnt=1; //dep是用来存每个点的深度
12 queue<int>r;
13 struct edge
14 {
15     int next,to,val,flag;
16 }q[M<<1],str1;  //要记得这里M<<1,我RE了好几次T_T
17 void add_edge(int u,int v,int w)  //存边（正、反）
18 {
19     q[cnt].next=fir[u];
20 
21     q[cnt].to=v;
22     q[cnt].val=w;
23     q[cnt].flag=cnt+1;
24     fir[u]=cnt++;
25     q[cnt].next=fir[v];
26 
27     q[cnt].to=u;
28     q[cnt].val=0;  //反向边的值应该设为0
29     q[cnt].flag=cnt-1; //这个flag的作用就是用来快速查找他的反向边
30     //我的上一个代码用vector来查找某个元素的时候太浪费时间了
31     fir[v]=cnt++;
32 }
33 int bfs() //每次先bfs查看有没有增广路
34 {
35     for(int i=1;i<=n;++i) dep[i]=0;
36     while(!r.empty()) r.pop();
37     r.push(a);
38     dep[a]=1;
39     while(!r.empty())
40     {
41         int now=r.front();
42         r.pop();
43         for(int i=fir[now];i;i=q[i].next)
44         {
45             int v=q[i].to;
46             if(q[i].val && !dep[v])
47             {
48                 dep[v]=dep[now]+1;
49                 r.push(v);
50             }
51         }
52     }
53     return dep[b];
54 }
55 int dfs(int now,int aim,int flow) //用来查找容量
56 {
57     if(now==aim || !flow)
58     {
59         return flow;
60     }
61     int res=0;
62     for(int i=fir[now];i;i=q[i].next)
63     {
64         int v=q[i].to;
65         if(q[i].val && dep[v]==dep[now]+1) //当容量不为0，而且这一个点是他的下一个深度
66         {
67             int kk=dfs(v,aim,min(flow,q[i].val));  //dfs找是否能到终点
68             res+=kk;  //最大流增加
69             flow-=kk; //可用流量减少
70             q[i].val-=kk; //正向边减少流量，反向边增加
71             q[q[i].flag].val+=kk;
72         }
73     }
74     return res;
75 }
76 int Dinic()
77 {
78     int ans=0;
79     while(bfs())  //只要有增广路就能增量
80     {
81         //printf("**
");
82         ans+=dfs(a,b,INF);
83     }
84     return ans;
85 }
86 int main()
87 {
88     memset(fir,0,sizeof(fir));
89     scanf("%d%d%d%d",&n,&m,&a,&b);
90     while(m--)
91     {
92         int u,v,w;
93         scanf("%d%d%d",&u,&v,&w);
94         add_edge(u,v,w);
95     }
96     printf("%d
",Dinic());
97     return 0;
98 }

View Code

 

Do not sincere non-interference。
Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is starvae must get to B within least time, it‘s said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once.


So, under a good RP, starvae may have many chances to get to city B. But he don‘t know how many chances at most he can make a data with the girl he likes . Could you help starvae?

InputThe first line is an integer T indicating the case number.(1<=T<=65)
For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0<c<=1000)it means there is a road from a to b and it‘s distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads from a to b, they are different.

At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.
There may be some blank line between each case.OutputOutput a line with a integer, means the chances starvae can get at most.Sample Input

3
7 8
1 2 1
1 3 1
2 4 1
3 4 1
4 5 1
4 6 1
5 7 1
6 7 1
1 7

6 7
1 2 1
2 3 1
1 3 3
3 4 1
3 5 1
4 6 1
5 6 1
1 6

2 2
1 2 1
1 2 2
1 2

Sample Output

2
1
1
网易翻译不用看来，看也看不懂，这道题就是让你去求出来从起点到终点能有几条最短路，那么这道题就采用 最短路&&最大流
最短路：求有几条最短路，那肯定要先用最短路算法求出来最短长度
最大流：最大流是求从起点到终点最大流度，这样我们只需要把每个边得权值设为1，这样一旦一个边走过，那按照最大流这个边就不可以走，就可以把这道题转化为最短路
在确定最短路中的都是那几条边才到的终点，我们可以d[起点]+边长==d[终点] 来判断，只要满足这个就说明最起码这一点是最短的，因为最短路求出来的每一段都是最短的
最大流：https://blog.csdn.net/stevensonson/article/details/79177530
下面上代码：

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<queue>
  4 #include<vector>
  5 #include<algorithm>
  6 #include<iostream>
  7 #define N 1050
  8 #define M 100050
  9 using namespace std;
 10 const int INF=0x3f3f3f3f;
 11 int n,m,a,b,fir[N],dep[N],cnt,dis[N],d[N];
 12 queue<int>r;
 13 struct edge
 14 {
 15     int next,to,val,flag;
 16 } q[M<<1],str1;
 17 struct shudui1
 18 {
 19     int start,value;
 20     bool operator <(const shudui1 q)const
 21     {
 22         return value>q.value;
 23     }
 24 } str11;
 25 struct shudui2
 26 {
 27     int start,value;
 28 } str2;
 29 vector<shudui2>w[1005];
 30 priority_queue<shudui1>t;
 31 int bfs() //每次先bfs查看有没有增广路
 32 {
 33     for(int i=1;i<=n;++i) dep[i]=0;
 34     while(!r.empty()) r.pop();
 35     r.push(a);
 36     dep[a]=1;
 37     while(!r.empty())
 38     {
 39         int now=r.front();
 40         r.pop();
 41         for(int i=fir[now];i;i=q[i].next)
 42         {
 43             int v=q[i].to;
 44             if(q[i].val && !dep[v])
 45             {
 46                 dep[v]=dep[now]+1;
 47                 r.push(v);
 48             }
 49         }
 50     }
 51     return dep[b];
 52 }
 53 int dfs(int now,int aim,int flow) //用来查找容量
 54 {
 55     if(now==aim || !flow)
 56     {
 57         return flow;
 58     }
 59     int res=0;
 60     for(int i=fir[now];i;i=q[i].next)
 61     {
 62         int v=q[i].to;
 63         if(q[i].val && dep[v]==dep[now]+1) //当容量不为0，而且这一个点是他的下一个深度
 64         {
 65             int kk=dfs(v,aim,min(flow,q[i].val));  //dfs找是否能到终点
 66             res+=kk;  //最大流增加
 67             flow-=kk; //可用流量减少
 68             q[i].val-=kk; //正向边减少流量，反向边增加
 69             q[q[i].flag].val+=kk;
 70         }
 71     }
 72     return res;
 73 }
 74 int Dinic()
 75 {
 76     int ans=0;
 77     while(bfs())  //只要有增广路就能增量
 78     {
 79         //printf("**
");
 80         ans+=dfs(a,b,INF);
 81     }
 82     return ans;
 83 }
 84 void add_edge(int u,int v,int w)
 85 {
 86     q[cnt].next=fir[u];
 87 
 88     q[cnt].to=v;
 89     q[cnt].val=w;
 90     q[cnt].flag=cnt+1;
 91     fir[u]=cnt++;
 92     q[cnt].next=fir[v];
 93 
 94     q[cnt].to=u;
 95     q[cnt].val=0;
 96     q[cnt].flag=cnt-1;
 97     fir[v]=cnt++;
 98 }
 99 int main()
100 {
101     int tt;
102     scanf("%d",&tt);
103     while(tt--)
104     {
105         cnt=1;
106         memset(fir,0,sizeof(fir));
107         scanf("%d%d",&n,&m);
108         for(int i=1; i<=n; ++i)
109             w[i].clear();
110         while(m--)
111         {
112             int u,v,ww;
113             scanf("%d%d%d",&u,&v,&ww);
114             str2.start=v;
115             str2.value=ww;
116             w[u].push_back(str2);
117         }
118         scanf("%d%d",&a,&b);
119         memset(d,INF,sizeof(d));
120         memset(dis,0,sizeof(dis));
121         str11.start=a;
122         str11.value=0;
123         t.push(str11);
124         d[str11.start]=0;
125         while(!t.empty())
126         {
127             str11=t.top();
128             t.pop();
129             int x=str11.start;
130             if(dis[x]) continue;
131             dis[x]=1;
132             int len=w[x].size();
133             for(int i=0; i<len; ++i)
134             {
135                 str2=w[x][i];
136                 if(d[str2.start]>d[x]+str2.value)
137                 {
138                     str11.value=d[str2.start]=d[x]+str2.value;
139                     str11.start=str2.start;
140                     t.push(str11);
141                 }
142             }
143         }
144 //        for(int i=1; i<=n; ++i)
145 //            printf("%d ",d[i]);
146 //        printf("
");
147         for(int i=1; i<=n; ++i)
148         {
149             int len=w[i].size();
150             for(int j=0; j<len; ++j)
151             {
152                 str2=w[i][j];
153                 if(d[i]+str2.value==d[str2.start])
154                 {
155                     //printf("%d%d***
",i,str2.start);
156                     add_edge(i,str2.start,1);
157                 }
158             }
159         }
160         printf("%d
",Dinic());
161     }
162     return 0;
163 }

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