Tarjan + bfs HYSBZ 1179Atm

1179: [Apio2009]Atm

Time Limit: 15 Sec Memory Limit: 162 MB
Submit: 3250 Solved: 1346
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Description

Input

第一行包含两个整数N、M。N表示路口的个数，M表示道路条数。接下来M行，每行两个整数，这两个整数都在1到N之间，第i+1行的两个整数表示第i条道路的起点和终点的路口编号。接下来N行，每行一个整数，按顺序表示每个路口处的ATM机中的钱数。接下来一行包含两个整数S、P，S表示市中心的编号，也就是出发的路口。P表示酒吧数目。接下来的一行中有P个整数，表示P个有酒吧的路口的编号

Output

输出一个整数，表示Banditji从市中心开始到某个酒吧结束所能抢劫的最多的现金总数。

Sample Input

6 7
1 2
2 3
3 5
2 4
4 1
2 6
6 5
10
12
8
16
1 5
1 4
4
3
5
6

Sample Output

HINT

50%的输入保证N, M<=3000。所有的输入保证N, M<=500000。每个ATM机中可取的钱数为一个非负整数且不超过4000。输入数据保证你可以从市中心沿着Siruseri的单向的道路到达其中的至少一个酒吧。

来源： http://www.lydsy.com/JudgeOnline/problem.php?id=1179

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#include<cstdio>

#include<cstring>

#include<iostream>

#include<string>

#include<vector>

#include<queue>

#include<stack>

#include<set>

#include<algorithm>

using namespace std;

#define N 500001

stack<int>Sta;

queue<int>Que;

vector<int>Gra[N],newGra[N];

bool isBar[N],inStack[N];

int low[N],dfn[N],belong[N],money[N],newMoney[N],cnt,Time,start;

int maxMoney[N];

void Tarjan(int s)

{

dfn[s] = low[s] = ++Time;

Sta.push(s);

inStack[s] = true;

for(int i=0;i<Gra[s].size();i++)

{

int j = Gra[s][i];

if(dfn[j] == 0){

Tarjan(j);

low[s] = min(low[s], low[j]);

}

else if(inStack[j] == true){

low[s] = min(low[s], dfn[j]);

}

if(dfn[s] == low[s])

{

cnt ++;

while(!Sta.empty()){

int temp = Sta.top(); Sta.pop();

belong[temp] = cnt;

newMoney[cnt] += money[temp];

inStack[temp] = false;

if(temp == s) break;

}

int spfa()

{

int ans = 0;

Que.push(start);

memset(maxMoney,0,sizeof(maxMoney));

maxMoney[start] = newMoney[start];

while(!Que.empty())

{

int now = Que.front(); Que.pop();

for(int i = 0; i < newGra[now].size(); i++)

{

int j = newGra[now][i];

if(maxMoney[now] + newMoney[j] > maxMoney[j])

{

maxMoney[j] = maxMoney[now] + newMoney[j];

Que.push(j);

}

if(isBar[j]) ans = max(ans, maxMoney[j]);

}

return ans;

}

int main()

{

int n,m,a,b,s,p;

scanf("%d%d",&n,&m);

for (int i = 0; i < m; i++) {

/* code */

scanf("%d%d",&a, &b);

Gra[a].push_back(b);

}

for (int i = 1; i <= n; i++) {

/* code */

scanf("%d",&money[i]);

}

scanf("%d%d",&s,&p);

memset(inStack,false,sizeof(inStack));

memset(newMoney,0,sizeof(newMoney));

memset(isBar,false,sizeof(isBar));

memset(dfn,0,sizeof(dfn));

cnt = Time = 0;

for(int i = 1; i <= n; i++) if(dfn[i] == 0) Tarjan(i);

if(cnt == 1) {printf("%d",newMoney[cnt]);return 0;}

for(int i=0;i<p;i++)

{

scanf("%d", &a);

isBar[belong[a]] = true;

}

start = belong[s];

for(int i =1; i <= n; i ++)

{

for(int j = 0; j < Gra[i].size(); j++){

int k = Gra[i][j];

if(belong[i] != belong[k]){

newGra[belong[i]].push_back(belong[k]);

}

int ans = spfa();

printf("%d", ans);

}

来源： http://tool.oschina.net/highlight

看起来，STL和链式向前星相比，速度还是慢了不少的

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#include<cstdio>

#include<cstring>

#include<iostream>

#include<string>

#include<vector>

#include<queue>

#include<stack>

#include<set>

#include<algorithm>

using namespace std;

#define N 500001

stack<int>Sta;

queue<int>Que;

struct edge

{

int u,v,next;

}edge1[N],edge2[N];

bool isBar[N],inStack[N];

int low[N],dfn[N],belong[N],money[N],newMoney[N],cnt,Time,start;

int maxMoney[N],head[N],head2[N];

void add(int u, int v, int id)

{

edge1[id].u = u;

edge1[id].v = v;

edge1[id].next = head[u];

head[u] = id;

}

void add2(int u, int v, int id)

{

edge2[id].u = u;

edge2[id].v = v;

edge2[id].next = head2[u];

head2[u] = id;

}

void Tarjan(int s)

{

dfn[s] = low[s] = ++Time;

Sta.push(s);

inStack[s] = true;

for(int u = head[s]; ~u; u = edge1[u].next)

{

int v = edge1[u].v;

if(dfn[v] == 0){

Tarjan(v);

low[s] = min(low[s], low[v]);

}

else if(inStack[v] == true){

low[s] = min(low[s], dfn[v]);

}

if(dfn[s] == low[s])

{

cnt ++;

while(!Sta.empty()){

int temp = Sta.top(); Sta.pop();

belong[temp] = cnt;

newMoney[cnt] += money[temp];

inStack[temp] = false;

if(temp == s) break;

}

int spfa()

{

int ans = 0;

Que.push(start);

memset(maxMoney,0,sizeof(maxMoney));

maxMoney[start] = newMoney[start];

while(!Que.empty())

{

int now = Que.front(); Que.pop();

for(int u = head2[now]; ~u; u = edge2[u].next)

{

int v = edge2[u].v;

if(maxMoney[now] + newMoney[v] > maxMoney[v])

{

maxMoney[v] = maxMoney[now] + newMoney[v];

Que.push(v);

}

if(isBar[v]) ans = max(ans, maxMoney[v]);

}

return ans;

}

int main()

{

int n,m,a,b,s,p;

scanf("%d%d",&n,&m);

memset(head,-1,sizeof(head));

for (int i = 0; i < m; i++) {

/* code */

scanf("%d%d",&a, &b);

add(a,b,i+1);

}

for (int i = 1; i <= n; i++) {

/* code */

scanf("%d",&money[i]);

}

scanf("%d%d",&s,&p);

memset(inStack,false,sizeof(inStack));

memset(newMoney,0,sizeof(newMoney));

memset(isBar,false,sizeof(isBar));

memset(dfn,0,sizeof(dfn));

cnt = Time = 0;

for(int i = 1; i <= n; i++) if(dfn[i] == 0) Tarjan(i);

if(cnt == 1) {printf("%d",newMoney[cnt]);return 0;}

for(int i=0;i<p;i++)

{

scanf("%d", &a);

isBar[belong[a]] = true;

}

start = belong[s];

memset(head2,-1,sizeof(head2));

int kkk = 0;

for(int i =1; i <= n; i ++)

{

for(int u = head[i]; ~u; u = edge1[u].next){

int v = edge1[u].v;

if(belong[i] != belong[v]){

add2(belong[i],belong[v],++kkk);

}

int ans = spfa();

printf("%d", ans);

}