HDU 6201 transaction transaction transaction（树上dfs/费用流）

Link：

http://acm.hdu.edu.cn/showproblem.php?pid=6201

题意：

有n个城市，每个城市有一本书，买书需要花钱，卖书可以得到钱，并且两个城市之间的距离也有花费。选择两个城市，一个城市买书一个城市卖书，问得到的最大收益。保证给定的n-1条边组成一棵树

解法：

这道题开了好久好久……一直想套费用流结果一直TLE，最后一小时才开始用dfs来写。。衰QAQ
先考虑树上每个点手上有书时的最小花费：这个书可能是在这一点买的，也可能是上一个点带过来的，因此自下往上从叶子结点开始更新，每次递归返回时更新父结点的cost。但是因为这样更新后叶子结点的兄弟还没有更新，所以再从上往下用父结点更新叶子结点，最后拿value-cost即可。
对于样例

3
8 3 16
1 2 2
1 3 1
　8
2/ \1
/ 　\
3 　16
第一次更新后 cost[1] = 5 cost[2] = 3 cost[3] = 16
第二次更新后 cost[1] = 5 cost[2] = 3 cost[3] = 6

代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <cmath>using namespace std;const int N = 1e5+10;const int inf = 1<<30;typedef pair<int,int> p;int vis[N];vector<p> child[N];int value[N];int cost[N];int ans = 0;void dfs1(int x) {    vis[x] = 1;    cost[x] = value[x];    for(int i = 0; i <child[x].size(); i++) {        if(!vis[child[x][i].first]) {            dfs1(child[x][i].first);            cost[x] = min(cost[x],cost[child[x][i].first]+child[x][i].second);        }    }}void dfs2(int x) {    vis[x] = 1;    for(int i = 0; i < child[x].size(); i++) {        if(!vis[child[x][i].first] ) {            if( cost[x] + child[x][i].second < cost[child[x][i].first] ){                cost[child[x][i].first] = cost[x] + child[x][i].second;            }            dfs2(child[x][i].first);        }    }    ans = max(ans,value[x] - cost[x]);    return;}int main() {//  freopen("1.txt","r",stdin);    int t;    scanf("%d", &t);    while(t--) {        ans = 0;        int n;        scanf("%d", &n);        for(int i = 1; i <= n; i++) {            child[i].clear();            vis[i] = 0;        }        int x, y ,z;        for(int i = 1; i <= n; i++) {            scanf("%d", &x);            value[i] = x;        }        for(int i = 1; i < n; i++) {            scanf("%d %d %d", &x, &y, &z);            child[x].push_back(p(y,z));            child[y].push_back(p(x,z));        }        dfs1(1);        memset(vis,0,sizeof(vis));        dfs2(1);        printf("%d\n",ans);    }    return 0;}/*238 3 161 2 21 3 1*/

最小费用流解法：
用边限制起点流量之后直接跑费用流，建图的话每个点x连源点s x’连汇点t x’连x。
x到y拆成x→y’ y→x’。
一定要起点流量限制！！！
边数是10n！！！

#include <iostream>#include <queue>#include <stack>#include <vector>#include <set>#include <cmath>#include <map>#include <cstdio>#include <cstring>using namespace std;typedef long long ll;  long long  _read()    {        char ch;      bool flag = false;      int a = 0;        while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));        if(ch != '-')      {          a *= 10;          a += ch - '0';        }      else      {          flag = true;      }      while(((ch = getchar()) >= '0') && (ch <= '9'))      {          a *= 10;          a += ch - '0';      }         if(flag)      {          a = -a;      }      return a;    }    const int MAXN = 200010;const int MAXM = 1200010;const int INF = 0x3f3f3f3f;struct Edge {    int to,next,cap,flow,cost;} edge[MAXM];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N;//节点总个数，节点编号从0~N-1void init(int n){    N = n;    tol = 0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int cap,int cost){    edge[tol].to = v;    edge[tol].cap = cap;    edge[tol].cost = cost;    edge[tol].flow = 0;    edge[tol].next = head[u];    head[u] = tol++;    edge[tol].to = u;    edge[tol].cap = 0;    edge[tol].cost = -cost;    edge[tol].flow = 0;    edge[tol].next = head[v];    head[v] = tol++;}bool spfa(int s,int t){    queue<int>q;    for(int i = 0; i < N; i++) {        dis[i] = INF;        vis[i] = false;        pre[i] = -1;    }    dis[s] = 0;    vis[s] = true;    q.push(s);    while(!q.empty()) {        int u = q.front();        q.pop();        vis[u] = false;        for(int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].to;            if(edge[i].cap > edge[i].flow &&                    dis[v] > dis[u] + edge[i].cost ) {                dis[v] = dis[u] + edge[i].cost;                pre[v] = i;                if(!vis[v]) {                    vis[v] = true;                    q.push(v);                }            }        }    }    if(pre[t] == -1)return false;    else return true;}//返回的是最大流，cost存的是最小费用int minCostMaxflow(int s,int t,int &cost,int n){    int flow = 0;    cost = 0;    while(spfa(s,t)) {         if(flow>1) break;        int Min = INF;        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {            if(Min > edge[i].cap - edge[i].flow)                Min = edge[i].cap - edge[i].flow;        }        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {            edge[i].flow += Min;            edge[i^1].flow -= Min;            cost += edge[i].cost * Min;        }        flow += Min;//        cout<<flow<<endl;    }    return flow;}int main(){//  freopen("1.txt","r",stdin);    int t;    t= _read();    while(t--){        int n;        n =_read();        int S = 0,s = 2*n+1, t = 2 * n + 2;        init(n*2+3);        int x, y ,z;        for(int i = 1; i <= n; i++){            x = _read();            addedge(s,i,1,x);            addedge(i+n,t,1,-x);            addedge(i+n,i,1,0);        }        for(int i = 1; i < n; i++){            x=_read(),y=_read(),z=_read();            addedge(x,y+n,1,z);            addedge(y,x+n,1,z);        }        x = 0;        addedge(S,s,1,0);//      cout<<"===="<<endl;//      continue;        int ans = minCostMaxflow(S,t,x,n);         x = -x;         if(x<0) x=0;        printf("%d\n",x);    }    return 0;}