算典04_例题_05_UVA-512

Spreadsheet Tracking

题意

题解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
#define met(a, b) memset(a, b, sizeof(a));

struct node {
char opt[5];
int n, a[20];
int x1, y1, x2, y2;
}cmd[maxn];
int n, r, c, on, x, y;

void solve(int x, int y) {
printf("Cell data in (%d,%d) ", x, y);
for(int i = 0; i < n; ++i) {
int t = x, k = y;
if(!strcmp(cmd[i].opt, "IR")) {
for(int j = 0; j < cmd[i].n; ++j) {
if(cmd[i].a[j] <= t) ++x;
}
}
else if(!strcmp(cmd[i].opt, "IC")) {
for(int j = 0; j < cmd[i].n; ++j) {
if(cmd[i].a[j] <= k) ++y;
}
}
else if(!strcmp(cmd[i].opt, "DR")) {
for(int j = 0; j < cmd[i].n; ++j) {
if(cmd[i].a[j] < t) --x;
else if(cmd[i].a[j] == t) {x = 0; i = n; break;}
}
}
else if(!strcmp(cmd[i].opt, "DC")) {
for(int j = 0; j < cmd[i].n; ++j) {
if(cmd[i].a[j] < k) --y;
else if(cmd[i].a[j] == k) {y = 0; i = n; break;}
}
}
else {  //EX
if(x == cmd[i].x1 && y == cmd[i].y1) x = cmd[i].x2, y = cmd[i].y2;
else if(x == cmd[i].x2 && y == cmd[i].y2) x = cmd[i].x1, y = cmd[i].y1;
}
}
if(x <= 0 /*|| x > r*/ || y <= 0 /*|| y > c*/) printf("GONE\n");
else printf("moved to (%d,%d)\n", x, y);
}

int main() {
#ifdef _LOCAL
freopen("in.txt", "r", stdin);
#endif

int kase = 0; bool line = 0;
while(scanf("%d%d", &r, &c) == 2 && (r || c)) {
scanf("%d", &n);
for(int i = 0; i < n; ++i) {
scanf("%s", cmd[i].opt);
if(!strcmp(cmd[i].opt, "EX")) scanf("%d%d%d%d", &cmd[i].x1, &cmd[i].y1, &cmd[i].x2, &cmd[i].y2);
else {scanf("%d", &cmd[i].n); for(int j = 0; j < cmd[i].n; ++j) scanf("%d", &cmd[i].a[j]); }
}

if(line) printf("\n");
if(!line) line = 1;
printf("Spreadsheet #%d\n", ++kase);
scanf("%d", &on);
while(on--) {
scanf("%d%d", &x, &y);
solve(x, y);
}
}
return 0;
}