开封网站开发,自建商城网站,wordpress文件功能详解,山东网站建设和游戏开发的公司P2656 采蘑菇 题意#xff1a; 有n个点#xff0c;m个单向边#xff0c;每个边都有边权#xff0c;如果经过这个边#xff0c;可以获得其边权#xff0c;而其边权会变成原来的p倍(0.1p0.8)#xff0c;向下取整 从s点出发#xff0c;问最多可以采到的蘑菇 题…P2656 采蘑菇 题意 有n个点m个单向边每个边都有边权如果经过这个边可以获得其边权而其边权会变成原来的p倍(0.1p0.8)向下取整 从s点出发问最多可以采到的蘑菇 题解 因为是单向边除非出现一个环不然每个边最多只能走一次如果有一个环环上的边权可以一直获取直到边权为0. 所以我们可以用tarjan进行缩点将这个环上所有得到的价值加起来赋给缩成的点x。 缩完点后就同时有点权(在之前环上所能获取的价值)和边权且无环那直接跑一个拓扑dp就可以在拓扑过程中转移dp,然后取最大即可 总结tarjan缩点拓扑dp 代码 // Problem: P2656 采蘑菇 // Contest: Luogu // URL: https://www.luogu.com.cn/problem/P2656 // Memory Limit: 125 MB // Time Limit: 1000 ms // By Jozky#include bits/stdc.h #include unordered_map #define debug(a, b) printf(%s %d\n, a, b); using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pairint, int PII; clock_t startTime, endTime; //Fe~Jozky const ll INF_ll 1e18; const int INF_int 0x3f3f3f3f; void read(){}; template typename _Tp, typename... _Tps void read(_Tp x, _Tps... Ar) {x 0;char c getchar();bool flag 0;while (c 0 || c 9)flag| (c -), c getchar();while (c 0 c 9)x (x 3) (x 1) (c ^ 48), c getchar();if (flag)x -x;read(Ar...); } template typename T inline void write(T x) {if (x 0) {x ~(x - 1);putchar(-);}if (x 9)write(x / 10);putchar(x % 10 0); } void rd_test() { #ifdef ONLINE_JUDGE #elsestartTime clock();freopen(data.in, r, stdin); #endif } void Time_test() { #ifdef ONLINE_JUDGE #elseendTime clock();printf(\nRun Time:%lfs\n, (double)(endTime - startTime) / CLOCKS_PER_SEC); #endif } const int maxn 3e5 9; double hui[maxn]; int n, m; struct node{int v,w;double k; }; vectornodevec[maxn]; int dfn[maxn], low[maxn], vis[maxn]; int color[maxn]; int tot 0, col 0; int degree[maxn]; vectorPII vec2[maxn]; vectorint co[maxn]; int newa[maxn]; stackint s; void tarjan(int x) {vis[x] 1;dfn[x] low[x] tot;s.push(x);for (auto it : vec[x]) {int v it.v;int w it.w;if (!dfn[v]) {tarjan(v);low[x] min(low[x], low[v]);}else if (vis[v]) {low[x] min(low[x], dfn[v]);}}if (dfn[x] low[x]) {col;while (1) {int top s.top();s.pop();color[top] col;vis[top] 0;if (top x)break;}} } int dp[maxn]; void solve() {for (int i 1; i n; i) {if (!dfn[i])tarjan(i);}for (int i 1; i n; i) {for (auto it : vec[i]) {int v it.v;int w it.w;double kit.k;if (color[i] ! color[v]) {degree[color[v]];vec2[color[i]].push_back({color[v], w});}else {while(w){newa[color[v]]w;ww*k/10;}}}} } void troop(int s) {queueint q;for(int i1;icol;i){if(!degree[i])q.push(i);dp[i]-INF_int;}dp[color[s]]newa[color[s]];q.push(s);while (!q.empty()) {int u q.front();q.pop();for (auto it : vec2[u]) {int v it.first;int w it.second;degree[v]--;dp[v] max(dp[v], dp[u] newa[v] w);if (!degree[v])q.push(v);}} } int main() {rd_test();read(n, m);for (int i 1; i m; i) {int u, v, w;double s;scanf(%d%d%d%lf, u, v, w, s);vec[u].push_back({v, w,s*10});}int s;read(s);solve();troop(s);int maxx 0;for (int i 1; i col; i)maxx max(dp[i], maxx);cout maxx;//Time_test(); } //92