POJ 3585Accumulation Degree

Posted 2020-08-17 ve-2021

【原题题面】传送门

【题解大意】

乍一看感觉以为网络流表示一点都不会不会不会。

然后发现可以树形dp搞一下。

再然后知道了换根dp。

设d[x]表示以x为根的子树中把x做为源点，从x出发流向子树的流量最大是多少。

d[x] += min(d[y],z);(deg[y]!=1)

d[x] = z; (deg[y]==1)

f[x]表示以x做为源点，从x出发流向整个水系的流量最大是多少。

f[y] = d[y] + min(f[x] - min(d[y],z));(deg[x]!=1)

f[y] = d[y] + z;

【code】

暴力：

#include<bits/stdc++.h>
using namespace std;
#define File "degree"
#define ll long long
inline void file(){
    freopen(File".in","r",stdin);
    freopen(File".out","w",stdout);
}
inline int read(){
    int x = 0,f = 1; char ch = getchar();
    while(ch < ‘0‘||ch > ‘9‘){if(ch == ‘-‘)f = -1; ch = getchar();}
    while(ch >= ‘0‘ && ch <= ‘9‘){x = (x<<1) + (x<<3) + ch-‘0‘; ch = getchar();}
    return x * f;
}
const int mxn = 2e5+3;
int n;
struct edge{
    int nxt,y,v;
}e[mxn<<1];

int to[mxn],len;
inline void add(int xx,int yy,int vv){
    e[++len].nxt = to[xx];
    to[xx] = len;
    e[len].y = yy;
    e[len].v = vv;
}
bool v[mxn];
int d[mxn],deg[mxn];

inline void dp(int x){
    v[x] = 1;
    d[x] = 0;
    for(int i = to[x]; i;i = e[i].nxt){
        int y = e[i].y,z = e[i].v;
        if(v[y]) continue;
        dp(y);
        if(deg[y]==1) d[x] += z;
        else d[x] += min(d[y],z);
    }
}
int T;
int main(){
//    file();
    T = read();
    while(T--){
        n = read();
        len = 1;
        memset(v,0,sizeof v);
        memset(d,0,sizeof d);
        memset(to,0,sizeof to);
        memset(deg,0,sizeof deg);
        for(int i = 1;i < n; ++i){
            int x = read(),y = read(),z = read();
            add(x,y,z),add(y,x,z);
            deg[x]++,deg[y]++; 
        }
        int ans(0);
        for(int i = 1;i <= n; ++i){
            memset(d,0,sizeof d);
            memset(v,0,sizeof v); 
            dp(i);
            ans = max(ans,d[i]);
        }
        printf("%d\n",ans); 
    } 
    return 0;
}

view code

标程：

#include<bits/stdc++.h>
using namespace std;
#define File "degree"
#define ll long long
inline void file(){
    freopen(File".in","r",stdin);
    freopen(File".out","w",stdout);
}
inline int read(){
    int x = 0,f = 1; char ch = getchar();
    while(ch < ‘0‘||ch > ‘9‘){if(ch == ‘-‘)f = -1; ch = getchar();}
    while(ch >= ‘0‘ && ch <= ‘9‘){x = (x<<1) + (x<<3) + ch-‘0‘; ch = getchar();}
    return x * f;
}
const int mxn = 2e5+3;
int T,n;
struct edge{
    int nxt,y,v;
}e[mxn<<1];

int to[mxn],len;
inline void add(int xx,int yy,int vv){
    e[++len].nxt = to[xx];
    to[xx] = len;
    e[len].y = yy;
    e[len].v = vv;
}
/*
1
5
1 2 11
1 4 13
3 4 5
4 5 10
*/

bool v[mxn];
int d[mxn],deg[mxn],f[mxn];

inline void dp(int x){
    v[x] = 1;
    d[x] = 0;
    for(int i = to[x]; i;i = e[i].nxt){
        int y = e[i].y,z = e[i].v;
        if(v[y]) continue;
        dp(y);
        if(deg[y]==1) d[x] += z;
        else d[x] += min(d[y],z);
    }
}

inline void dfs(int x){
    v[x] = 1;
    for(int i = to[x]; i;i = e[i].nxt){
        int y = e[i].y,z = e[i].v;
        if(v[y]) continue;
        if(deg[x]==1) f[y] = d[y] + z;
        else f[y] = d[y] + min(f[x] - min(d[y],z),z);
        dfs(y);
    }
}

int main(){
//    file();
    T = read();
    while(T--){ 
        n = read();
        len = 1;
        memset(v,0,sizeof v);
        memset(f,0,sizeof f);
        memset(d,0,sizeof d);
        memset(deg,0,sizeof deg);
        memset(to,0,sizeof to);
        for(int i = 1;i < n; ++i){
            int x = read(),y = read(),z = read();
            add(x,y,z),add(y,x,z);
            deg[x]++,deg[y]++;
        }
        int rt(1);
        dp(rt);
        memset(v,0,sizeof v);
        f[rt] = d[rt];
        dfs(rt);
        int ans(0);
        for(int i = 1;i <= n; ++i) ans = max(ans,f[i]);
        printf("%d\n",ans);
    } 
    return 0;
}

View Code