2015 Multi-University Training Contest 1 Tricks Device
Tricks Device
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connecting the chambers. Innocent Wu wants to catch up Dumb Zhang to find out the answers of some questions, however, it’s Dumb Zhang’s intention to keep Innocent Wu in the dark, to do which he has to stop Innocent Wu from getting him. Only via the original shortest ways from the entrance to the end of the tomb costs the minimum time, and that’s the only chance Innocent Wu can catch Dumb Zhang.
Unfortunately, Dumb Zhang masters the art of becoming invisible(奇门遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.
Unfortunately, Dumb Zhang masters the art of becoming invisible(奇门遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.
Input
There are multiple test cases. Please process till EOF.
For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.
In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0<li<=100) minute to pass channel i.
The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.
For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.
In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0<li<=100) minute to pass channel i.
The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.
Output
Output two numbers to stand for the answers of Dumb Zhang and Innocent Wu’s questions.
Sample Input
8 9
1 2 2
2 3 2
2 4 1
3 5 3
4 5 4
5 8 1
1 6 2
6 7 5
7 8 1
Sample Output
2 6
解题:最短路+最小割
先把所有的最短路提取到另一份图中,然后看看最少经过几条边(可以用dp优化,或者标记已经访问的边来加速)可以由终点到起点
m-减去最少的可经过的边 即可删除的边
然后再对刚才提取的图 求最小割,最小割即为最少删除几条边,可以使得最短路变长
需要注意重边的影响
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 #define pii pair<int,int> 3 using namespace std; 4 const int maxn = 1000010; 5 const int INF = 0x3f3f3f3f; 6 struct arc { 7 int to,w,next,id; 8 arc(int x = 0,int y = 0,int z = -1) { 9 to = x; 10 w = y; 11 next = z; 12 } 13 } e[maxn]; 14 int d[maxn],tot,S,T,head[maxn],cur[maxn]; 15 vector< pii >g[maxn]; 16 void add(int u,int v,int wa,int wb,int id = 0) { 17 e[tot] = arc(v,wa,head[u]); 18 e[tot].id = id; 19 head[u] = tot++; 20 e[tot] = arc(u,wb,head[v]); 21 e[tot].id = id; 22 head[v] = tot++; 23 } 24 bool done[maxn]; 25 priority_queue< pii,vector< pii >,greater< pii > >q; 26 void dijkstra() { 27 while(!q.empty()) q.pop(); 28 memset(d,0x3f,sizeof d); 29 d[S] = 0; 30 memset(done,false,sizeof done); 31 q.push(pii(d[S],S)); 32 while(!q.empty()) { 33 int u = q.top().second; 34 q.pop(); 35 if(done[u]) continue; 36 done[u] = true; 37 for(int i = head[u]; ~i; i = e[i].next) { 38 if(d[e[i].to] > d[u] + e[i].w) { 39 d[e[i].to] = d[u] + e[i].w; 40 g[e[i].to].clear(); 41 g[e[i].to].push_back(pii(u,e[i].id)); 42 q.push(pii(d[e[i].to],e[i].to)); 43 } else if(d[e[i].to] == d[u]+e[i].w) { 44 g[e[i].to].push_back(pii(u,e[i].id)); 45 q.push(pii(d[e[i].to],e[i].to)); 46 } 47 } 48 } 49 } 50 int minstep; 51 void dfs(int u,int dep,int fa) { 52 if(u == S) { 53 minstep = min(dep,minstep); 54 return; 55 } 56 for(int i = g[u].size()-1; i >= 0; --i) { 57 if(g[u][i].first == fa) continue; 58 dfs(g[u][i].first,dep+1,u); 59 bool flag = true; 60 for(int j = head[g[u][i].first]; flag && ~j; j = e[j].next) { 61 if(e[j].id == g[u][i].second) flag = false; 62 } 63 if(flag) { 64 add(g[u][i].first,u,1,0,g[u][i].second); 65 //cout<<g[u][i]<<" *** "<<u<<endl; 66 } 67 } 68 } 69 bool bfs() { 70 queue<int>q; 71 memset(d,-1,sizeof d); 72 d[S] = 1; 73 q.push(S); 74 while(!q.empty()) { 75 int u = q.front(); 76 q.pop(); 77 for(int i = head[u]; ~i; i = e[i].next) { 78 if(e[i].w && d[e[i].to] == -1) { 79 d[e[i].to] = d[u] + 1; 80 q.push(e[i].to); 81 } 82 } 83 } 84 return d[T] > -1; 85 } 86 int dfs(int u,int low) { 87 if(u == T) return low; 88 int tmp = 0,a; 89 for(int &i = cur[u]; ~i; i = e[i].next) { 90 if(e[i].w &&d[e[i].to] == d[u]+1&&(a=dfs(e[i].to,min(e[i].w,low)))) { 91 e[i].w -= a; 92 e[i^1].w += a; 93 low -= a; 94 tmp += a; 95 if(!low) break; 96 } 97 } 98 if(!tmp) d[u] = -1; 99 return tmp; 100 } 101 int dinic() { 102 int ret = 0; 103 while(bfs()) { 104 memcpy(cur,head,sizeof head); 105 ret += dfs(S,INF); 106 } 107 return ret; 108 } 109 int main() { 110 int n,m,u,v,w; 111 while(~scanf("%d%d",&n,&m)) { 112 for(int i = tot = 0; i < maxn; ++i) { 113 g[i].clear(); 114 head[i] = -1; 115 } 116 for(int i = 0; i < m; ++i) { 117 scanf("%d%d%d",&u,&v,&w); 118 add(u,v,w,w,i); 119 } 120 S = 1; 121 T = n; 122 dijkstra(); 123 minstep = INT_MAX; 124 memset(head,-1,sizeof head); 125 tot = 0; 126 dfs(T,0,-1); 127 int by = m-minstep; 128 int ax = dinic(); 129 printf("%d %d\n",ax,by); 130 } 131 return 0; 132 }
重新写了下,思路更清楚些
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 using namespace std; 3 using PII = pair<int,int>; 4 const int maxn = 2010; 5 const int INF = 0x3f3f3f3f; 6 struct arc { 7 int to,w,next; 8 arc(int x = 0,int y = 0,int z = -1) { 9 to = x; 10 w = y; 11 next = z; 12 } 13 } e[1000100]; 14 int head[maxn],hd[maxn],tot,S,T,n,m; 15 int gap[maxn],d[maxn]; 16 bool done[maxn]; 17 void add(int head[maxn],int u,int v,int wa,int wb) { 18 e[tot] = arc(v,wa,head[u]); 19 head[u] = tot++; 20 e[tot] = arc(u,wb,head[v]); 21 head[v] = tot++; 22 } 23 void dijkstra() { 24 memset(d,0x3f,sizeof d); 25 memset(done,false,sizeof done); 26 priority_queue<PII,vector<PII>,greater<PII>>q; 27 d[S] = 0; 28 q.push(PII(0,S)); 29 while(!q.empty()) { 30 int u = q.top().second; 31 q.pop(); 32 if(done[u]) continue; 33 done[u] = true; 34 for(int i = hd[u]; ~i; i = e[i].next) { 35 if(d[e[i].to] > d[u] + e[i].w) { 36 d[e[i].to] = d[u] + e[i].w; 37 q.push(PII(d[e[i].to],e[i].to)); 38 } 39 } 40 } 41 } 42 void build() { 43 for(int i = 1; i <= n; ++i) { 44 for(int j = hd[i]; ~j; j = e[j].next) { 45 if(d[e[j].to] == d[i] + e[j].w) 46 add(head,i,e[j].to,1,0); 47 } 48 } 49 } 50 int bfs() { 51 memset(gap,0,sizeof gap); 52 memset(d,-1,sizeof d); 53 queue<int>q; 54 d[T] = 0; 55 q.push(T); 56 while(!q.empty()) { 57 int u = q.front(); 58 q.pop(); 59 ++gap[d[u]]; 60 for(int i = head[u]; ~i; i = e[i].next) { 61 if(d[e[i].to] == -1) { 62 d[e[i].to] = d[u] + 1; 63 q.push(e[i].to); 64 } 65 } 66 } 67 return d[S]; 68 } 69 int dfs(int u,int low) { 70 if(u == T) return low; 71 int tmp = 0,minH = n - 1; 72 for(int i = head[u]; ~i; i = e[i].next) { 73 if(e[i].w) { 74 if(d[e[i].to] + 1 == d[u]) { 75 int a = dfs(e[i].to,min(e[i].w,low)); 76 e[i].w -= a; 77 e[i^1].w += a; 78 tmp += a; 79 low -= a; 80 if(!low) break; 81 if(d[S] >= n) return tmp; 82 } 83 if(e[i].w) minH = min(minH,d[e[i].to]); 84 } 85 } 86 if(!tmp) { 87 if(--gap[d[u]] == 0) d[S] = n; 88 ++gap[d[u] = minH + 1]; 89 } 90 return tmp; 91 } 92 int sap(int ret = 0) { 93 while(d[S] < n) ret += dfs(S,INF); 94 return ret; 95 } 96 int main() { 97 int u,v,w; 98 while(~scanf("%d%d",&n,&m)) { 99 memset(head,-1,sizeof head); 100 memset(hd,-1,sizeof hd); 101 for(int i = tot = 0; i < m; ++i) { 102 scanf("%d%d%d",&u,&v,&w); 103 add(hd,u,v,w,w); 104 } 105 S = 1; 106 T = n; 107 dijkstra(); 108 build(); 109 int y = m - bfs(),x = sap(); 110 printf("%d %d\n",x,y); 111 } 112 return 0; 113 }
SPFA貌似更快些,这图稀疏
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <bits/stdc++.h> 2 using namespace std; 3 using PII = pair<int,int>; 4 const int maxn = 2010; 5 const int INF = 0x3f3f3f3f; 6 struct arc { 7 int to,w,next; 8 arc(int x = 0,int y = 0,int z = -1) { 9 to = x; 10 w = y; 11 next = z; 12 } 13 } e[1000100]; 14 int head[maxn],hd[maxn],tot,S,T,n,m; 15 int gap[maxn],d[maxn]; 16 bool in[maxn] = {}; 17 void add(int head[maxn],int u,int v,int wa,int wb) { 18 e[tot] = arc(v,wa,head[u]); 19 head[u] = tot++; 20 e[tot] = arc(u,wb,head[v]); 21 head[v] = tot++; 22 } 23 void dijkstra() { 24 memset(d,0x3f,sizeof d); 25 queue<int>q; 26 d[S] = 0; 27 q.push(S); 28 while(!q.empty()){ 29 int u = q.front(); 30 q.pop(); 31 in[u] = false; 32 for(int i = hd[u]; ~i; i = e[i].next){ 33 if(d[e[i].to] > d[u] + e[i].w){ 34 d[e[i].to] = d[u] + e[i].w; 35 if(!in[e[i].to]){ 36 in[e[i].to] = true; 37 q.push(e[i].to); 38 } 39 } 40 } 41 } 42 } 43 void build() { 44 for(int i = 1; i <= n; ++i) { 45 for(int j = hd[i]; ~j; j = e[j].next) { 46 if(d[e[j].to] == d[i] + e[j].w) 47 add(head,i,e[j].to,1,0); 48 } 49 } 50 } 51 int bfs() { 52 memset(gap,0,sizeof gap); 53 memset(d,-1,sizeof d); 54 queue<int>q; 55 d[T] = 0; 56 q.push(T); 57 while(!q.empty()) { 58 int u = q.front(); 59 q.pop(); 60 ++gap[d[u]]; 61 for(int i = head[u]; ~i; i = e[i].next) { 62 if(d[e[i].to] == -1) { 63 d[e[i].to] = d[u] + 1; 64 q.push(e[i].to); 65 } 66 } 67 } 68 return d[S]; 69 } 70 int dfs(int u,int low) { 71 if(u == T) return low; 72 int tmp = 0,minH = n - 1; 73 for(int i = head[u]; ~i; i = e[i].next) { 74 if(e[i].w) { 75 if(d[e[i].to] + 1 == d[u]) { 76 int a = dfs(e[i].to,min(e[i].w,low)); 77 e[i].w -= a; 78 e[i^1].w += a; 79 tmp += a; 80 low -= a; 81 if(!low) break; 82 if(d[S] >= n) return tmp; 83 } 84 if(e[i].w) minH = min(minH,d[e[i].to]); 85 } 86 } 87 if(!tmp) { 88 if(--gap[d[u]] == 0) d[S] = n; 89 ++gap[d[u] = minH + 1]; 90 } 91 return tmp; 92 } 93 int sap(int ret = 0) { 94 while(d[S] < n) ret += dfs(S,INF); 95 return ret; 96 } 97 int main() { 98 int u,v,w; 99 while(~scanf("%d%d",&n,&m)) { 100 memset(head,-1,sizeof head); 101 memset(hd,-1,sizeof hd); 102 for(int i = tot = 0; i < m; ++i) { 103 scanf("%d%d%d",&u,&v,&w); 104 add(hd,u,v,w,w); 105 } 106 S = 1; 107 T = n; 108 dijkstra(); 109 build(); 110 int y = m - bfs(),x = sap(); 111 printf("%d %d\n",x,y); 112 } 113 return 0; 114 }
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