UVa 12186 Another Crisis dp：树上dp

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UVA - 12186

Another Crisis

Time Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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The question is from here.

My Solution

d(i)表示给上级发信至少需要多少工人。把所有子结点的d值排序然后取前c个，

对除非的处理 ★★ c = (k*T - 1) /100 +1;

用前向星储存树每组数据搞完要 for 1 -> n 对sons[ i ].clear();

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;const int maxn = 100008;int n,t,T;vector<int> sons[maxn];      //sons[i] 为结点i的子列表int dp(int u){    if(sons[u].empty()) return 1;    int k = sons[u].size();    vector<int> d;    for(int i = 0; i < k; i++)        d.push_back(dp(sons[u][i]));    sort(d.begin(),d.end());    int c = (k*T-1) /100 +1;    int ans = 0;    for(int i = 0; i < c; i++) ans += d[i];    return ans;}int main(){    while(scanf("%d%d", &n, &T)){        if(n == 0 && T == 0) break;        for(int i = 1; i <= n; i++){            scanf("%d", &t);            sons[t].push_back(i);        }        printf("%d\n", dp(0));        for(int i = 0; i <= n; i++){            sons[i].clear();        }    }    return 0;}

也可以写成这样

<span style="font-size:14px;">#include <iostream>#include <cstdio>#include <vector>#include <algorithm>//#define LOCALusing namespace std;const int maxn = 100008;int n,t,T;vector<int> sons[maxn];      //sons[i] 为结点i的子列表void build(){    for(int i = 1; i <= n; i++){        scanf("%d", &t);        sons[t].push_back(i);    }}int dp(int u){    if(sons[u].empty()) return 1;    int k = sons[u].size();    vector<int> d;    for(int i = 0; i < k; i++)        d.push_back(dp(sons[u][i]));    sort(d.begin(),d.end());    int c = (k*T-1) /100 +1;    int ans = 0;    for(int i = 0; i < c; i++) ans += d[i];    return ans;}void initial(){    for(int i = 0; i <= n; i++){   //清空的时候要把大boss的也清理掉        sons[i].clear();    }}int main(){    #ifdef LOCAL    freopen("a.txt","r",stdin);    #endif // LOCAL    while(scanf("%d%d", &n, &T)){        if(n == 0 && T == 0) break;        build();        printf("%d\n", dp(0));        initial();    }    return 0;}</span>

Thank you！

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