标签:scanf 最大化 can efi out 结束 stack 分享 img
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define rap(i, a, n) for(int i=a; i<=n; i++) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 10010, INF = 0x7fffffff; int ID[maxn], IN[maxn], vis[maxn], pre[maxn]; struct node { int u, v, c, b; }Node[maxn*2]; bool dirmst(int root, int n, int m, int cost, int B) { LL ans = 0; while(true) { for(int i=0; i<n; i++) IN[i] = INF; //记录最小前驱边的值 //1、找最小前驱边 for(int i=0; i<m; i++) { int u = Node[i].u; int v = Node[i].v; if(Node[i].c < IN[v] && u != v && Node[i].b >= B) { pre[v] = u; IN[v] = Node[i].c; // cout<< e.v << " " << e.u <<endl; } } //2、判断是否联通 for(int i=0; i<n; i++) { if(i == root) continue; if(IN[i] == INF) return false; } //3、找环 int cntnode = 0; mem(ID, -1); mem(vis, -1); IN[root] = 0; for(int i=0; i<n; i++) { ans += IN[i]; int v = i; while(vis[v] != i && ID[v] == -1 && v != root) { vis[v] = i; v = pre[v]; } //如果存在环 则把环中的点缩为一个点 if(v != root && ID[v] == -1) { for(int j=pre[v]; j!=v; j=pre[j]) { ID[j] = cntnode; } ID[v] = cntnode++; } } if(cntnode == 0) break; //没有环就结束 //重新标记其它点 for(int i=0; i<n; i++) if(ID[i] == -1) ID[i] = cntnode++; for(int i=0; i<m; i++) { int v = Node[i].v; Node[i].u = ID[Node[i].u]; Node[i].v = ID[Node[i].v]; if(Node[i].u != Node[i].v) Node[i].c -= IN[v]; } n = cntnode; root = ID[root]; } if(ans <= cost) return true; return false; } int main() { int T, n, m, cost; scanf("%d", &T); while(T--) { int l = INF, r = 0; scanf("%d%d%d", &n, &m, &cost); for(int i=0; i<m; i++) { scanf("%d%d%d%d", &Node[i].u, &Node[i].v, &Node[i].b, &Node[i].c); Node[i+m] = Node[i]; l = min(l, Node[i].b); r = max(r, Node[i].b); } if(!dirmst(0, n, m, cost, l)) { printf("streaming not possible.\n"); continue; } while(l <= r) { int mid = l + (r - l) / 2; for(int i=0; i<m; i++) Node[i] = Node[i+m]; if(!dirmst(0, n, m, cost, mid)) r = mid - 1; else l = mid + 1; } printf("%d kbps\n", r); } return 0; }
Stream My Contest UVA - 11865(带权最小树形图+二分最小值最大化)
标签:scanf 最大化 can efi out 结束 stack 分享 img
原文地址:https://www.cnblogs.com/WTSRUVF/p/9426490.html