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.

 

 

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.

 

 

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

 

 

  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 }

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