UVa 185 - Roman Numerals

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題目：給你一個羅馬數組組成的計算式A+B=C，判斷等式是否成立，有兩種方案：

1.羅馬數字：每個字母代表一個數字，相鄰數字左邊的小就右邊減去左邊的數字，否則加和；

2.阿拉伯數字：每個字母用不同的數字代替（無前導零）。

分析：圖論，搜索。數據較小，利用回溯法搜索即可，可以按數字搜索，也可以按字母搜索。

說明：用char存儲數據溢出，好久才發現，╮(╯▽╰)╭。

#include <cstring>#include <cstdio>char str[33], buf[3][11], maps[9];int  roman[128], first[128], visit[128], used[10], digit[128];int  ans;int  value[3];void dfs(int d, int n){if (ans > 1) return;if (d >= n) {for (int i = 0; i < 3; ++ i) {value[i] = 0;for (int j = 0; buf[i][j]; ++ j)value[i] = value[i]*10 + digit[buf[i][j]];}ans += (value[0]+value[1] == value[2]);return;}else {for (int i = 0; i <= 9; ++ i) {if (used[i] || !i && first[maps[d]]) continue;digit[maps[d]] = i;used[i] = 1;dfs(d+1, n);used[i] = 0;if (ans > 1) return;}}}int main(){roman['I'] = 1; roman['V'] = 5;roman['X'] = 10;roman['L'] = 50;roman['C'] = 100;roman['D'] = 500;roman['M'] = 1000;int rvalue[3];while (gets(str) && str[0] != '#') {int size = 0, count = 0, sum = 0;memset(visit, 0, sizeof(visit));for (int i = 0; str[i]; ++ i) {if (str[i] == '+' || str[i] == '=') {buf[size ++][count] = 0;count = 0;}else {buf[size][count ++] = str[i];if (!visit[str[i]]) {visit[str[i]] = 1;maps[sum ++] = str[i];}}}buf[size ++][count] = 0;memset(first, 0, sizeof(first));for (int i = 0; i < 3; ++ i) {rvalue[i] = 0;for (int j = 0; buf[i][j]; ++ j) {if (buf[i][j+1] && roman[buf[i][j]] < roman[buf[i][j+1]])rvalue[i] -= roman[buf[i][j]];else rvalue[i] += roman[buf[i][j]];}first[buf[i][0]] = 1;}if (rvalue[0]+rvalue[1] == rvalue[2])printf("Correct ");else printf("Incorrect ");ans = 0;memset(used, 0, sizeof(used));dfs(0, sum);if (ans > 1)printf("ambiguous\n");else if (ans == 1)printf("valid\n");else printf("impossible\n");}    return 0;}

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