Poj 3592 Instantaneous Transference

人生的第二个强连通~

每个点有一个权值，求缩点后，从起点出发的任意一条路径中权值最大的一条

坑是如果一个点传送到图外的位置，这个边也要建起来

具体见代码

----------------我是代码的分割线------------------

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<stack>
#include<cctype>
using namespace std;

const int maxm = 45;
const int maxn = 45*45;

vector<int> edge[maxn];
vector<int> Scc[maxn];

stack<int> S;

int dfn[maxn],low[maxn],_cnt,bel[maxn];

void init(int n){
for(int i=0;i<=n;i++){
edge[i].clear();
Scc[i].clear();
}
while(S.empty()==false)
S.pop();
memset(dfn,-1,sizeof(dfn));
memset(low,-1,sizeof(low));
_cnt = 1;
memset(bel,-1,sizeof(bel));
}

#define iter vector<int>::iterator

void dfs(int st){
S.push(st);
dfn[st] = low[st] = _cnt ++;
for(iter it = edge[st].begin();it!=edge[st].end();it++){
int x = *it;
if(dfn[x] == -1){
// printf("-1 %d %d\n",st,x);
dfs(x);
low[st] = min(low[st],low[x]);
}
else if(bel[x] == -1){
// printf(" %d %d n = %d\n",st,x,dfn[x]);
low[st] = min(low[st],dfn[x]);
// printf("low = %d\n",low[st]);
}
}
if(low[st] == dfn[st]){
while(S.top()!=st){
bel[S.top()] = st;
S.pop();
}
bel[st] = st;
S.pop();
}
}

char Map[maxm][maxm];
int n,m;

int toID(int i,int j){
return i*m+j;
}

void Link(int i,int j,int p,int q){
if(Map[p][q] != '#'){
// printf("%d %d\n",toID(i,j),toID(p,q));
// printf("%d - %d %d - %d\n",i,j,p,q);
edge[toID(i,j)].push_back(toID(p,q));
}
}

void initMap(){
for(int i=0;i<n-1;i++){
for(int j=0;j<m;j++){
Link(i,j,i+1,j);
}
}
for(int i=0;i<n;i++){
for(int j=0;j<m-1;j++){
Link(i,j,i,j+1);
}
}
}

int val[maxn];

int dp[maxn];

int cal(int st){
if(dp[st] != -1)
return dp[st];
int ma = 0;
for(iter it = Scc[st].begin();it != Scc[st].end();it++){
ma = max(ma,cal(*it));
}
return dp[st] = ma + val[st];
}

void out(char *p,int *s,int n){
printf("%s\t",p);
for(int i=0;i<n;i++)
printf(i<n-1?"%d ":"%d\n",s[i]);
}

#define debug(a) out(#a,a,n*m);

int main(){
int T;
scanf("%d",&T);
while(T-- && ~scanf("%d %d",&n,&m)){
for(int i=0;i<n;i++)
scanf("%s",Map[i]);
init(n*m);
initMap();
int p,q;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(Map[i][j] == '*'){
scanf("%d %d",&p,&q);
Link(i,j,p,q);
}
}
}
memset(val,0,sizeof(val));
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
int x = 0;
char c = Map[i][j];
if('0'<=c && c<='9')
x = c - '0';
val[toID(i,j)] = x;
}
}
for(int i=0;i<m*n;i++)
if(dfn[i]==-1)
dfs(i);
// debug(dfn);
// debug(low);
// debug(bel);
for(int i=0;i<n*m;i++){
if(bel[i] != i){
val[bel[i]] += val[i];
val[i] = 0;
}
}
for(int i=0;i<n*m;i++){
for(iter it = edge[i].begin();it!=edge[i].end();it++){
int x = *it;
if(bel[x] != bel[i]){
Scc[bel[i]].push_back(bel[x]);
}
}
}
memset(dp,-1,sizeof(dp));
printf("%d\n",cal(bel[0]));
}
return 0;
}


感觉这种缩点的题有有种莫名的精神愉悦感Orz