题意:

  一个人在一张大图上走,给你路径与起点,求他走出的矩形面积并.(大概这个意思自行百度标题...

SOL:

  与其说这是一道图论题不如说是一道生动活泼的STL-vector教学....

  离散化宽搜,没什么别的...vector用得淋漓尽致...

Code:

  

/*==========================================================================
# Last modified: 2016-03-18 08:32
# Filename: t1.cpp
# Description:
==========================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <ctime>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm> #include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <deque> #define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1]
#define pb push_back
#define find(a,b) lower_bound((a).begin(), (a).end(), (b))-(a).begin() #define INF 10000000000
#define maxn 3000
#define maxm 100000
#define pi 3.1415926535898
#define _e 2.718281828459
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template<class T> inline
void read(T& num) {
bool start=false,neg=false;
char c;
num=0;
while((c=getchar())!=EOF) {
if(c=='-') start=neg=true;
else if(c>='0' && c<='9') {
start=true;
num=num*10+c-'0';
} else if(start) break;
}
if(neg) num=-num;
}
/*==================split line==================*/
ll n, x[maxn], x1 = INF/2, y1 = INF/2, ans = 0;
char d[maxn];
vector<ll> x2,y2;
int a[maxn][maxn];
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1}; void check(){
FORP(i,0,x2.size()-1) printf("%lld ",x2[i]);
printf("\n");
FORP(i,0,y2.size()-1) printf("%lld ",y2[i]);
cout << endl;
}
void mark(int ld,int rd,int lu,int ru) {
if (ld>lu) swap(ld,lu); if (rd>ru) swap(rd,ru);
for (int i=ld;i<=lu;i++)
for (int j=rd;j<=ru;j++) a[i][j] = 1;
}
queue<int> qx,qy;
void bfs() {
qx.push(0); qy.push(0);
while (!qx.empty()){
int nx=qx.front(),ny=qy.front();
qx.pop(); qy.pop();
FORP(i,0,3) {
int xx=nx+dx[i],yy=ny+dy[i];
if (xx<0||xx>=x2.size()||yy<0||yy>=y2.size()||a[xx][yy])
continue;
else {
a[xx][yy]=2;
qx.push(xx); qy.push(yy);
}
}
}
//FORP(i,0,3) dfs(xx+dx[i],yy+dy[i]);
}
void init(){
read(n);
x2.pb(0); x2.pb(INF);
y2.pb(0); y2.pb(INF);
FORP(i,0,n-1){
x2.pb(x1); x2.pb(x1+1);
y2.pb(y1); y2.pb(y1+1);
cin >> d[i]; read(x[i]);
if (d[i]=='R')x1+=x[i];if (d[i]=='L')x1-=x[i];
if (d[i]=='U')y1+=x[i];if (d[i]=='D')y1-=x[i];
}
x2.pb(x1); x2.pb(x1+1);
y2.pb(y1); y2.pb(y1+1);
// check();
sort(x2.begin(), x2.end()); sort(y2.begin(), y2.end());
x2.erase(unique(x2.begin(),x2.end()),x2.end());
y2.erase(unique(y2.begin(),y2.end()),y2.end()); // check();
}
int main(){
//freopen("a.in","r",stdin);
init(); int xnow=find(x2,INF/2),ynow=find(y2,INF/2);
ll xx=INF/2, yy=INF/2;
FORP(i,0,n-1){
int x4=xnow,y4=ynow;
if (d[i]=='R') xx+=x[i];if (d[i]=='L') xx-=x[i];
if (d[i]=='U') yy+=x[i];if (d[i]=='D') yy-=x[i];
xnow=find(x2,xx); ynow=find(y2,yy);
mark(xnow, ynow, x4, y4);
}
bfs(); FORP(i,0,x2.size()-1)
FORP(j,0,y2.size()-1)
if (a[i][j]!=2) ans+=(x2[i+1]-x2[i])*(y2[j+1]-y2[j]);
printf("%lld\n",ans);
return 0;
}
04-22 18:42