The main idea: give a map, and the value of n points (after buying this point, the point of departure from the point will be dyed)
Ask at least how much to spend to buy some, in order to make all the points of this picture is dyed
If not all points are dyed, the output cannot be dyed at the minimum value of the point
Problem-solving ideas: First Find all the strong connected components, and then find the minimum value of all the points within each strong connected component, because the point within the same strong connected component only need to buy one can be fully dyed
Then the point is shrunk, connected by a bridge, to form a new picture (the point below is the indentation of the strongly connected component)
Find all points with an entry level of 0, because points with a degree of 0 are not dyed at any other point, so they can only buy at points with a degree of 0
#include <cstdio>#include <cstring>#define N 3010#define M 8010#define INF 0x3f3f3f3f#define min (a) (a) < (b)? (a): (b))structedge{int from, to, next;} E[M];intN, Tot, p, M, Dfs_clock, scc_cnt, top;intHead[n], Cost[n], pre[n], sccno[n], stack[n], Lowlink[n],inch[N],value[N];voidAddedge (intUintV) {e[tot].to = v; E[tot]. from= u; E[tot].next = Head[u]; Head[u] = tot++;}voidInit () {memset (cost,0x3f,sizeof(cost)); scanf"%d", &p);intNo, Money; for(inti =0; I < P; i++) {scanf ("%d%d", &no, &money); Cost[no] = money; } scanf ("%d", &m); Memset (Head,-1,sizeof(head)); tot =0;intU, v; for(inti =0; I < m; i++) {scanf ("%d%d", &u, &v); Addedge (U, v); }}voidDfsintu) {pre[u] = lowlink[u] = ++dfs_clock; Stack[++top] = u;intV for(inti = Head[u]; I! =-1; i = e[i].next) {v = e[i].to;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++; while(1) {v = stack[top--]; SCCNO[V] = scc_cnt;if(v = = u) Break; } }}voidSolve () {memset (PRE,0,sizeof(pre)); memset (Sccno,0,sizeof(SCCNO)); Dfs_clock = scc_cnt = top =0; for(inti =1; I <= N; i++)if(!pre[i]) DFS (i); for(inti =1; I <= scc_cnt; i++) {value[I] = INF;inch[I] =1; } for(inti =1; I <= N; i++) {value[Sccno[i]] = min (value[Sccno[i]], cost[i]); }intU, V, MiniD; for(inti =0; i < tot; i++) {u = sccno[e[i]. from]; v = sccno[e[i].to];if(U! = V) {inch[V] =0; } }BOOLFlag =false; for(inti =1; I <= N; i++) {u = sccno[i];if(inch[U] &&value[u] = = INF) {MiniD = i; Flag =true; Break; } }if(flag) {printf ("no\n%d\n", MiniD);return; }intAns =0; for(inti =1; I <= scc_cnt; i++) {if(inch[i]) ans + =value[i]; } printf ("yes\n%d\n", ans);}intMain () { while(SCANF ("%d", &n)! = EOF) {init (); Solve (); }return 0;}
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FZU-1719 Spy Network (strong connectivity component)