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("**\n"); 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("**\n"); 99 printf("%d\n",Dinic()); 100 return 0; 101 }
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("**\n"); 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\n",Dinic()); 97 return 0; 98 }
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?
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("**\n"); 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("\n"); 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***\n",i,str2.start); 156 add_edge(i,str2.start,1); 157 } 158 } 159 } 160 printf("%d\n",Dinic()); 161 } 162 return 0; 163 }