HDU 5934

来源：互联网发布：淘宝上的官方折扣店编辑：程序博客网时间：2024/06/28 10:37

2016CCPC杭州赛，3题铁，B题当时出不了，太水了。

对每一块强连通分量(SCC)缩点，记录每一个点的入度，对入度为0的点的耗费求和为结果。

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <cstdlib>#include <stack>using namespace std;const int maxn = 1005;vector<int> G[maxn];int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_cnt;int d[maxn], cost[maxn];stack<int> S;struct Node{    int x, y, r, c;}e[maxn];void dfs(int u){    pre[u] = lowlink[u] = ++dfs_clock;    S.push(u);    for(int i = 0; i < G[u].size(); ++i)    {        int v = G[u][i];        if(!pre[v])        {            dfs(v);            lowlink[u] = min(lowlink[u], lowlink[v]);        }        else if(!sccno[v])        {            lowlink[u] = min(lowlink[u], pre[v]);        }    }    if(lowlink[u] == pre[u])    {        ++scc_cnt;        for(;;)        {            int x = S.top();S.pop();            sccno[x] = scc_cnt;            if(x == u)                break;        }    }}void find_scc(int n){    dfs_clock = scc_cnt = 0;    memset(sccno, 0, sizeof(sccno));    memset(pre, 0, sizeof(pre));    for(int i = 0; i < n; ++i)        if(!pre[i])            dfs(i);}int main(){    int kase = 0;    int T;    scanf("%d", &T);    while(T--)    {        memset(d, 0, sizeof(d));        memset(cost, 0x3f3f3f3f, sizeof(cost));        int n;        scanf("%d", &n);        for(int i = 0; i < n; ++i)        {            G[i].clear();            scanf("%d%d%d%d", &e[i].x, &e[i].y, &e[i].r, &e[i].c);            for(int j = 0; j < i; ++j)            {                long long x = (long long)e[i].x - (long long)e[j].x;                long long y = (long long)e[i].y - (long long)e[j].y;                long long d = x * x + y * y;                long long d1 = (long long)e[i].r * (long long)e[i].r;                long long d2 = (long long)e[j].r * (long long)e[j].r;                if(d1 >= d)                {                    G[i].push_back(j);                }                if(d2 >= d)                {                    G[j].push_back(i);                }            }        }        find_scc(n);        for(int i = 0; i < n; ++i)        {//            cout << "i == " << sccno[i] << endl;            int u = sccno[i];            cost[u] = min(cost[u], e[i].c);            for(int j = 0; j < G[i].size(); ++j)            {                int x = G[i][j];                int v = sccno[x];                if(u != v)                    ++d[v];                cost[v] = min(cost[v], e[x].c);            }        }        int ans = 0;        for(int i = 1; i <= scc_cnt; ++i)        {            if(d[i] == 0)            {                ans += cost[i];            }        }        printf("Case #%d: %d\n", ++kase, ans);//    }}

0 0